2023.03.14 17:44 *MoreTransportation79* **I need help. Which of these are gonna be used for graphing?**

submitted by MoreTransportation79 to askmath [link] [comments]

2023.01.15 12:03 *saberline152* **[University math]limits**

how would one solve the following limits?

lim for x ->0 ( (1/arctanh(x) ) - ( 1/(e^x-1) ) )

lim for x ->0 ( (1/arccosh(x) ) - ( 1/(e^x-1) ) )

I tried substitution with the identities but it still gets me nowhere, according to my Nspire, the first one has 1/2 as a solution and the second one has -infinity as a solution. I also tried looking for answers using Symbolab and how it works it out but it cannot work it out for the first one.

submitted by saberline152 to learnmath [link] [comments]
lim for x ->0 ( (1/arctanh(x) ) - ( 1/(e^x-1) ) )

lim for x ->0 ( (1/arccosh(x) ) - ( 1/(e^x-1) ) )

I tried substitution with the identities but it still gets me nowhere, according to my Nspire, the first one has 1/2 as a solution and the second one has -infinity as a solution. I also tried looking for answers using Symbolab and how it works it out but it cannot work it out for the first one.

2022.10.14 14:28 *drede_knig* **Help with limit values and absolutes**

Hey there! My teacher has passed out an assignment including a problem that I have been somewhat stuck on:

Apologies for the formatting, I haven't quite figured out how to use the fancy pants editor that reddit has.

*should* be correct and continuous. But I have never used absolutes before when working with limit values, so I figured there was tomfoolery at hand.

The work I've done:

However, what it suggests is weird, and I disagree with it. Symbolab wants the limit value approaching from below to mean that x = -x. Its reasoning being that:

I've messed up before because convoluted setting like this trip me up, or because I've overlooked simple interpretations of rules like this. So I'm inclined to blindly trust the calculator on this one, which I don't want to do before checking with someone more knowledgeable than me first. Is it correct that if x is a negative value then x = -x?

submitted by drede_knig to MathHelp [link] [comments]
Apologies for the formatting, I haven't quite figured out how to use the fancy pants editor that reddit has.

Is f(x) continuous in x = 0 in the following? f(x) {(2x^2 + x)/x : x != 0, 1 : x = 0}So I have come to the conclusion that this

The work I've done:

First check for f(x) = (2x^2 + x)/x: I choose to approach f(0) with limit values from both sides to see if they converge. lim x -> 0- (all values approaching 0 from below are negative so x is negative, represented by -x) f(x) = (2x^2 + x)/-x : -x = x f(x) = (2x^2 + x)/x f(x) = 2x + 1 f(0) = 2*0 + 1 f(0) = 1 lim x lim x -> 0+ (all values approaching 0 from above are positive so x is positive,) f(x) = (2x^2 + x)/x : x = x And this is the same as above f(0) = 1 f(0) = 1 applies to both functions above, therefore the function is continuous in x = 0Usually I would've left it at this, but since we've never been given assignments including absolutes and limit values, nor could I find anything in my textbook about it, I felt uncertain. So I hopped onto Symbolab to check my work, and lo and behold, it disagrees with me.

However, what it suggests is weird, and I disagree with it. Symbolab wants the limit value approaching from below to mean that x = -x. Its reasoning being that:

x -> 0- means that x is negative, therefore x = -xI can see how it would think this. If x is a negative number, e.g (-2), then x = -(-2), meaning that x is the negative of x. However I struggle to agree with a value coming out of an absolute with a negation in front of it.

I've messed up before because convoluted setting like this trip me up, or because I've overlooked simple interpretations of rules like this. So I'm inclined to blindly trust the calculator on this one, which I don't want to do before checking with someone more knowledgeable than me first. Is it correct that if x is a negative value then x = -x?

2022.05.14 10:22 *oobeing* **Multivariable limit, stuck at last step of evaluation**

Limit is

lim (x,y) (0,0) x^{2}y^{2}/(x^{2}+y^{4})

I've tried looking if it DNE by going along x and y-axis etc, get 0 on all of them.Changed to polar coordinates and then simplified I get

lim r 0 r^{2}*cos^{2}x*sin^{2}x/(cos^{2}x+r^{2}sin^{4}x)

Then if you plug in you get

0/cos^{2}x

and symbolab says the answer here is 0, because 0/a = 0 when a =/= 0.But how can one know the angle x is not pi/2, making the denominator 0 as well?

Edit: Since limit must be independent upon which path you take for it to exist, meaning it must be independent of the angle x here, wouldn't this mean it doesn't exist since there's one angle/path for which the limit is not defined?

submitted by oobeing to learnmath [link] [comments]
lim (x,y) (0,0) x

I've tried looking if it DNE by going along x and y-axis etc, get 0 on all of them.Changed to polar coordinates and then simplified I get

lim r 0 r

Then if you plug in you get

0/cos

and symbolab says the answer here is 0, because 0/a = 0 when a =/= 0.But how can one know the angle x is not pi/2, making the denominator 0 as well?

Edit: Since limit must be independent upon which path you take for it to exist, meaning it must be independent of the angle x here, wouldn't this mean it doesn't exist since there's one angle/path for which the limit is not defined?

2021.02.13 01:31 *voidantis* **[Calculus: Limits] I have the answer but not sure how to get there.**

The question I'm stuck on is lim *x*→36 [(36 − *x) / (*6 − sqrt x)]. I know that the answer is 12. I used this limit calculator: https://www.symbolab.com/solvelimit-calculato%5Clim_%7Bx%5Cto36%7D%5Cfrac%7B%5Cleft(36-x%5Cright)%7D%7B6-%5Csqrt%7Bx%7D%7D%7D%7B6-%5Csqrt%7Bx%7D%7D) and I'm stuck on the step where it says rationalize the denominator. I just don't understand how we ended up with 6 + sqrt x. I understand plugging it in at the end and getting 12 there, just stuck on the middle step. Thank you to anyone who can explain this to me!

submitted by voidantis to HomeworkHelp [link] [comments]
2020.10.10 13:28 *Cuntsu* **question about limits**

Hi all the question I have is

lim x -> 0 ((csc(3x))(x+sin(x)))

I got the answer which is 2/3, but when I plugged it in a calculator(symbolab) it says that the limit diverges. However when I typed in the same question but changed csc(3x) into 1/sin(3x), it now says 2/3. Am I missing something? very much confused with this one.

I should also note that I used other online calculators (emathhelp, wolframalpha, etc) and they all say 2/3. Just confused if I did something wrong or its just symbolab.

any help will be appreciated thanks.

submitted by Cuntsu to MathHelp [link] [comments]
lim x -> 0 ((csc(3x))(x+sin(x)))

I got the answer which is 2/3, but when I plugged it in a calculator(symbolab) it says that the limit diverges. However when I typed in the same question but changed csc(3x) into 1/sin(3x), it now says 2/3. Am I missing something? very much confused with this one.

I should also note that I used other online calculators (emathhelp, wolframalpha, etc) and they all say 2/3. Just confused if I did something wrong or its just symbolab.

any help will be appreciated thanks.

2020.02.04 04:29 *MethaMathemine* **A Question for a Possibly Riemann Sum Problem**

*The characters are copied and pasted from *Symbolab*.

Question: Evaluate \lim _{n\to \:\infty }\sum _{k=1}^n\:\:\left(\frac{n}{k^2+n^2}\right)

I assume the problem uses 'Riemann Sum', because the integral uses two different variables -*n* and *k -* that appears to be characteristic of sample questions of the said concept.

From my understanding, the Riemann's Sum states that if*f* is continuous and integrable on [a,b], then

\int f\left(x\right)dx = \lim _{n\to \:\infty }\sum _{i=0}^{\infty }f\left(x_i\right)\Delta x

For this particular question, I've expressed it as follows:

i) \lim _{n\to \:\infty }\sum _{k=1}^n\:\:\left(\frac{n}{k^2+n^2}\right)

ii) \int _a^b\:f\left(x\right)dx\:\rightarrow \:\lim _{n\to \infty }\sum _{i=1}^nf\left(x_i\right)\Delta x

I choose an interval on [0,1], then

iii) Delta x = (b - a) / n \:\rightarrow \: 1/n

∴

iv) \:\lim _{n\to \infty }\sum _{i=1}^nf\left(x_i\right)\Delta x \rightarrow \lim _{n\to \infty }\sum _{i=1}^n\left(\frac{1}{n}\right)\left(\frac{n^2}{k^2+\:n^2}\:\right)

From here I'm stuck, and do not know what further steps are required to progress.

submitted by MethaMathemine to learnmath [link] [comments]
Question: Evaluate \lim _{n\to \:\infty }\sum _{k=1}^n\:\:\left(\frac{n}{k^2+n^2}\right)

I assume the problem uses 'Riemann Sum', because the integral uses two different variables -

From my understanding, the Riemann's Sum states that if

\int f\left(x\right)dx = \lim _{n\to \:\infty }\sum _{i=0}^{\infty }f\left(x_i\right)\Delta x

For this particular question, I've expressed it as follows:

i) \lim _{n\to \:\infty }\sum _{k=1}^n\:\:\left(\frac{n}{k^2+n^2}\right)

ii) \int _a^b\:f\left(x\right)dx\:\rightarrow \:\lim _{n\to \infty }\sum _{i=1}^nf\left(x_i\right)\Delta x

I choose an interval on [0,1], then

iii) Delta x = (b - a) / n \:\rightarrow \: 1/n

∴

iv) \:\lim _{n\to \infty }\sum _{i=1}^nf\left(x_i\right)\Delta x \rightarrow \lim _{n\to \infty }\sum _{i=1}^n\left(\frac{1}{n}\right)\left(\frac{n^2}{k^2+\:n^2}\:\right)

From here I'm stuck, and do not know what further steps are required to progress.

2020.01.31 00:07 *worldopp* **Lim f(x)=x^.5*sin(1/x)**

f(x)=x^.5*sin(1/x)

Lim x-> o f(x) from the positive side is zero per the squeeze theorem

Lim x-> o f(x) from the negative side cannot exist, as the domain of the function is x>0

So am I correct in saying that the Lim x-> o f(x) doesn't exist because Lim x-> o f(x) from the positive side DOESN'T EQUAL Lim x-> o f(x) from the negative side? If so, why does SymboSolver state that the limit DOES exist: https://www.symbolab.com/solvequadratic-equation-calculato%5Clim_%7Bx%5Cto0%7D%5Cleft(x%5E%7B.5%7Dsin%5Cleft(%5Cfrac%7B1%7D%7Bx%7D%5Cright)%5Cright)%5Cright))

submitted by worldopp to MathHelp [link] [comments]
Lim x-> o f(x) from the positive side is zero per the squeeze theorem

Lim x-> o f(x) from the negative side cannot exist, as the domain of the function is x>0

So am I correct in saying that the Lim x-> o f(x) doesn't exist because Lim x-> o f(x) from the positive side DOESN'T EQUAL Lim x-> o f(x) from the negative side? If so, why does SymboSolver state that the limit DOES exist: https://www.symbolab.com/solvequadratic-equation-calculato%5Clim_%7Bx%5Cto0%7D%5Cleft(x%5E%7B.5%7Dsin%5Cleft(%5Cfrac%7B1%7D%7Bx%7D%5Cright)%5Cright)%5Cright))

2019.10.01 03:05 *IndependentRaccoon5* **Discrepency between finding a multivariable limit using polar coordinates vs using paths**

Hello everyone,

So I'm in vector calculus right now and we re learning how to compute limits of multivariable fncs. I know that if we approach from different curves and the resulting limit is different, then limit does not exist. I did that with this fnc and found that it doesn't exist. Explanation in symbolab link below.

https://www.symbolab.com/solvemulti-var-limit-calculato%5Clim_%7B%5Cleft(x%2C%20y%5Cright)%5Cto%5Cleft(0%2C%200%5Cright)%7D%5Cleft(%5Cfrac%7Bsin%5Cleft(2x%5Cright)-2x%2By%7D%7Bx%5E%7B3%7D%2By%7D%5Cright)%5Cto%5Cleft(0%2C%200%5Cright)%7D%5Cleft(%5Cfrac%7Bsin%5Cleft(2x%5Cright)-2x%2By%7D%7Bx%5E%7B3%7D%2By%7D%5Cright))

However, when my friend attempted this problem by substituting polar coordinates into the previous fnc, he got the lim as r approaches 0 , the limit approaches 1. We confirmed his answer on symbol lab and wolfram alpha and both algorithms say his answer is correct. I have attached the symbolab for this below.

https://www.symbolab.com/solvestep-by-step/%5Clim_%7Br%5Cto0%7D%5Cleft(%5Cfrac%7Bsin%5Cleft(2rcos%5Ctheta%5Cright)-2rcos%5Ctheta%2Brsin%5Ctheta%7D%7Br%5E%7B3%7Dcos%5E%7B3%7D%5Ctheta%2Brsin%5Ctheta%7D%5Cright)-2rcos%5Ctheta%2Brsin%5Ctheta%7D%7Br%5E%7B3%7Dcos%5E%7B3%7D%5Ctheta%2Brsin%5Ctheta%7D%5Cright))

If anyone could help me point out why there is a discrepancy when using polar coordinates, I would be super grateful

submitted by IndependentRaccoon5 to learnmath [link] [comments]
So I'm in vector calculus right now and we re learning how to compute limits of multivariable fncs. I know that if we approach from different curves and the resulting limit is different, then limit does not exist. I did that with this fnc and found that it doesn't exist. Explanation in symbolab link below.

https://www.symbolab.com/solvemulti-var-limit-calculato%5Clim_%7B%5Cleft(x%2C%20y%5Cright)%5Cto%5Cleft(0%2C%200%5Cright)%7D%5Cleft(%5Cfrac%7Bsin%5Cleft(2x%5Cright)-2x%2By%7D%7Bx%5E%7B3%7D%2By%7D%5Cright)%5Cto%5Cleft(0%2C%200%5Cright)%7D%5Cleft(%5Cfrac%7Bsin%5Cleft(2x%5Cright)-2x%2By%7D%7Bx%5E%7B3%7D%2By%7D%5Cright))

However, when my friend attempted this problem by substituting polar coordinates into the previous fnc, he got the lim as r approaches 0 , the limit approaches 1. We confirmed his answer on symbol lab and wolfram alpha and both algorithms say his answer is correct. I have attached the symbolab for this below.

https://www.symbolab.com/solvestep-by-step/%5Clim_%7Br%5Cto0%7D%5Cleft(%5Cfrac%7Bsin%5Cleft(2rcos%5Ctheta%5Cright)-2rcos%5Ctheta%2Brsin%5Ctheta%7D%7Br%5E%7B3%7Dcos%5E%7B3%7D%5Ctheta%2Brsin%5Ctheta%7D%5Cright)-2rcos%5Ctheta%2Brsin%5Ctheta%7D%7Br%5E%7B3%7Dcos%5E%7B3%7D%5Ctheta%2Brsin%5Ctheta%7D%5Cright))

If anyone could help me point out why there is a discrepancy when using polar coordinates, I would be super grateful

2019.10.01 02:27 *IndependentRaccoon5* **Computing multivariable limits by converting to polar coordinates**

Hello everyone,

So I'm in vector calculus right now and we re learning how to compute limits of multivariable fncs. I know that if we approach from different curves and the resulting limit is different, then limit does not exist. I did that with this fnc and found that it doesn't exist. Explanation in symbolab link below.

https://www.symbolab.com/solvemulti-var-limit-calculato%5Clim_%7B%5Cleft(x%2C%20y%5Cright)%5Cto%5Cleft(0%2C%200%5Cright)%7D%5Cleft(%5Cfrac%7Bsin%5Cleft(2x%5Cright)-2x%2By%7D%7Bx%5E%7B3%7D%2By%7D%5Cright)%5Cto%5Cleft(0%2C%200%5Cright)%7D%5Cleft(%5Cfrac%7Bsin%5Cleft(2x%5Cright)-2x%2By%7D%7Bx%5E%7B3%7D%2By%7D%5Cright))

However, when my friend attempted this problem by substituting polar coordinates into the previous fnc, he got the lim as r approaches 0 , the limit approaches 1. We confirmed his answer on symbol lab and wolfram alpha and both algorithms say his answer is correct. I have attached the symbolab for this below.

https://www.symbolab.com/solvestep-by-step/%5Clim_%7Br%5Cto0%7D%5Cleft(%5Cfrac%7Bsin%5Cleft(2rcos%5Ctheta%5Cright)-2rcos%5Ctheta%2Brsin%5Ctheta%7D%7Br%5E%7B3%7Dcos%5E%7B3%7D%5Ctheta%2Brsin%5Ctheta%7D%5Cright)-2rcos%5Ctheta%2Brsin%5Ctheta%7D%7Br%5E%7B3%7Dcos%5E%7B3%7D%5Ctheta%2Brsin%5Ctheta%7D%5Cright))

If anyone could help me point out why there is a discrepancy when using polar coordinates, I would be super grateful

Edit: included links

submitted by IndependentRaccoon5 to MathHelp [link] [comments]

So I'm in vector calculus right now and we re learning how to compute limits of multivariable fncs. I know that if we approach from different curves and the resulting limit is different, then limit does not exist. I did that with this fnc and found that it doesn't exist. Explanation in symbolab link below.

https://www.symbolab.com/solvemulti-var-limit-calculato%5Clim_%7B%5Cleft(x%2C%20y%5Cright)%5Cto%5Cleft(0%2C%200%5Cright)%7D%5Cleft(%5Cfrac%7Bsin%5Cleft(2x%5Cright)-2x%2By%7D%7Bx%5E%7B3%7D%2By%7D%5Cright)%5Cto%5Cleft(0%2C%200%5Cright)%7D%5Cleft(%5Cfrac%7Bsin%5Cleft(2x%5Cright)-2x%2By%7D%7Bx%5E%7B3%7D%2By%7D%5Cright))

However, when my friend attempted this problem by substituting polar coordinates into the previous fnc, he got the lim as r approaches 0 , the limit approaches 1. We confirmed his answer on symbol lab and wolfram alpha and both algorithms say his answer is correct. I have attached the symbolab for this below.

https://www.symbolab.com/solvestep-by-step/%5Clim_%7Br%5Cto0%7D%5Cleft(%5Cfrac%7Bsin%5Cleft(2rcos%5Ctheta%5Cright)-2rcos%5Ctheta%2Brsin%5Ctheta%7D%7Br%5E%7B3%7Dcos%5E%7B3%7D%5Ctheta%2Brsin%5Ctheta%7D%5Cright)-2rcos%5Ctheta%2Brsin%5Ctheta%7D%7Br%5E%7B3%7Dcos%5E%7B3%7D%5Ctheta%2Brsin%5Ctheta%7D%5Cright))

If anyone could help me point out why there is a discrepancy when using polar coordinates, I would be super grateful

Edit: included links

2019.09.25 07:25 *tenlegdragon* **Limit question with a denominator that gives ln0 by substitution.**

We're supposed to do this without graphs and without approaching from both sides numerically.

I don't want to post the question cause I'm not sure what university policy is on that since it's for marks (a minimal amount) even though I'm doing it at home and allowed to get help...

But I'll post something similar since I really just want to know how to approach the problem rather than the answer.

lim of f(x) as x tends to 5 of a fraction

Numerator: x^{x} raised x a couple more times. I guess it doesn't really matter because I can freely substitute x=5 and get the value which would just be a giant number

**Denominator:**

ln x - 5

I tried thinking of it as a composite function. g(x) = x-5 As x tends to 5, the limit is zero, but ln(0) is undefined, so when I go back to the equation I have 5^{5} all divided by "undefined"????

Where am I going wrong here? I put the graph in symbolab so I know I'm supposed to get -ve infinity, but what is the technique or order of steps/thoughts/tactics to get there?

Also, any links to somewhere where I can get info/practice on these "tricky" types of limit problems that can be solved without graphs, calculators or differentiation?

I guess a better way to put it is how do I approach: lim as x tends to a in f(x) = (exponential function)/ ln x-a or any situation where ln 0 comes up in a denominator situation where I can't just say "undefined"

submitted by tenlegdragon to MathHelp [link] [comments]
I don't want to post the question cause I'm not sure what university policy is on that since it's for marks (a minimal amount) even though I'm doing it at home and allowed to get help...

But I'll post something similar since I really just want to know how to approach the problem rather than the answer.

lim of f(x) as x tends to 5 of a fraction

Numerator: x

ln x - 5

I tried thinking of it as a composite function. g(x) = x-5 As x tends to 5, the limit is zero, but ln(0) is undefined, so when I go back to the equation I have 5

Where am I going wrong here? I put the graph in symbolab so I know I'm supposed to get -ve infinity, but what is the technique or order of steps/thoughts/tactics to get there?

Also, any links to somewhere where I can get info/practice on these "tricky" types of limit problems that can be solved without graphs, calculators or differentiation?

I guess a better way to put it is how do I approach: lim as x tends to a in f(x) = (exponential function)/ ln x-a or any situation where ln 0 comes up in a denominator situation where I can't just say "undefined"

2019.09.01 16:45 *hctiwte* **A limit using L'Hopital's rule**

So i have the limit:

lim(x->0) (2sinx - sin(2x)) / (sin(x) - xcos(x))

I have entered this limit in both Symbolab and Wolfram Alpha, they both solve it using L'Hopital's rule i can't see the steps because i don't have the pro version. Any help would be appreciated.

submitted by hctiwte to askmath [link] [comments]
lim(x->0) (2sinx - sin(2x)) / (sin(x) - xcos(x))

I have entered this limit in both Symbolab and Wolfram Alpha, they both solve it using L'Hopital's rule i can't see the steps because i don't have the pro version. Any help would be appreciated.

2019.03.20 12:19 *Urkedurke* **[High School Math] Limits - The limit of ln(x)/x^2 using L'hospital rule**

When I try and use the rule I get:

lim x->0+ ln(x)/x^{2} = 1/x/2x = 1/2x^{2} = ∞ , but if I plug it into symbolab I get -∞. Why?

submitted by Urkedurke to learnmath [link] [comments]
lim x->0+ ln(x)/x

2019.03.14 01:03 *LampGoat* **Find the limit WITHOUT L’Hôpital’s Rule**

I’m not allowed to use L’Hôpital’s rule yet but I can’t figure out how to do this question without it:

lim(x->0) 2xcscx

The only thing I can think of doing is: 2 * lim(x->0)xcscx 2 * lim(x->0)x/sinx 2 * lim(x->0)(sinx/x)^{-1} 2 * 1 =2

Using symbolab/L’Hôpital’s rule also results in 2 but I’m not sure if my method is correct. Edit: sorry for formatting on mobile

submitted by LampGoat to MathHelp [link] [comments]
lim(x->0) 2xcscx

The only thing I can think of doing is: 2 * lim(x->0)xcscx 2 * lim(x->0)x/sinx 2 * lim(x->0)(sinx/x)

Using symbolab/L’Hôpital’s rule also results in 2 but I’m not sure if my method is correct. Edit: sorry for formatting on mobile

2018.10.21 23:40 *OnePunchFan8* **Help with limits**

lim X->4 ((x+4)^0.5-(2X)^0.5)/(X-4))

How do I find the limit? It just seems to be undefined, as (4)-4=0. I tried using this site: https://www.symbolab.com/solvelimit-calculato%5Clim_%7Bx%5Cto4%7D%5Cleft(%5Cfrac%7B%5Csqrt%7Bx%2B4%7D-%5Csqrt%7B2X%7D%7D%7B%5Cleft(x-4%5Cright)%7D%5Cright%7D%5Cright))

, but I don't understand the step by step process.

I also need help with lim x->inf (4x^2-9)^0.5/(x-5), Calc here:

https://www.symbolab.com/solvelimit-calculato%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cleft(%5Cleft(%5Cfrac%7B%5Csqrt%7B4x%5E%7B2%7D-9%7D%7D%7Bx-5%7D%5Cright)%5Cright%5Cright))

submitted by OnePunchFan8 to calculus [link] [comments]
How do I find the limit? It just seems to be undefined, as (4)-4=0. I tried using this site: https://www.symbolab.com/solvelimit-calculato%5Clim_%7Bx%5Cto4%7D%5Cleft(%5Cfrac%7B%5Csqrt%7Bx%2B4%7D-%5Csqrt%7B2X%7D%7D%7B%5Cleft(x-4%5Cright)%7D%5Cright%7D%5Cright))

, but I don't understand the step by step process.

I also need help with lim x->inf (4x^2-9)^0.5/(x-5), Calc here:

https://www.symbolab.com/solvelimit-calculato%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cleft(%5Cleft(%5Cfrac%7B%5Csqrt%7B4x%5E%7B2%7D-9%7D%7D%7Bx-5%7D%5Cright)%5Cright%5Cright))

2018.10.01 22:28 *achunkypid* **Help understanding Trig Limits (calculus)**

Hey guys the book I'm reading is not very good at telling me exactly what I need to do for problems such as these below : lim x-> 0 sinx/x^{2} -x

and

lim x-> x+tanx/sinx

Symbolab keeps telling me to use L'hopitals rule but we have not been taught that yet nor is it taught in the section of the book we are on..

https://www3.canyons.edu/faculty/lel/math211/211-EX2-PART2-F16.pdf

The problems are on page 3.

submitted by achunkypid to cheatatmathhomework [link] [comments]
and

lim x-> x+tanx/sinx

Symbolab keeps telling me to use L'hopitals rule but we have not been taught that yet nor is it taught in the section of the book we are on..

https://www3.canyons.edu/faculty/lel/math211/211-EX2-PART2-F16.pdf

The problems are on page 3.

2018.03.03 02:26 *RLMarina* **Limits at 0**

I'm looking over an old question I did can't understand how I came to the answer. symbolab and mathway tell me to use l'Hopital rule but I didn't do that to reach my answer.

The problem is: lim x0 sinx/tan7x

the answer is 1/7

submitted by RLMarina to cheatatmathhomework [link] [comments]
The problem is: lim x0 sinx/tan7x

the answer is 1/7

2018.01.18 12:29 *AttDominate* **convergence of arrays - the criterium of d'Alembert. Not sure if doing it right.**

I apologize for my english and propably using wrong terms. English is not my native language.

I have to check if array is convergent or not using the criterium of d'Alembert. One of the problems is as fallows : [\; \Sigma_{n=1}^{{\infty}} ;] [\; {2n+5}\over{(n+4)! e^{n}} ;]

After computing [\; {n}\over{n+1} ;] i get [\; {(2n+5)(n+5)e}\over{2n+7} ;] which i checked on Wolfram Alpha and symbolab and both say it's correct.

Now I have to plug it into [\; \lim ;] if I understand correctly. So: [\; \lim_{n->\infty} ;] [\; {(2n+5)(n+5)e}\over{2n+7} ;] = [\; \lim_{n->\infty} ;] [\; {2n}\over{2e} ;]

I know this is perfectly valid solution. But i have 3 similar examples which all end up going to [\; \infty ;] and that's why I have doubts if im doing it right.

submitted by AttDominate to learnmath [link] [comments]
I have to check if array is convergent or not using the criterium of d'Alembert. One of the problems is as fallows : [\; \Sigma_{n=1}

After computing [\; {n}\over{n+1} ;] i get [\; {(2n+5)(n+5)e}\over{2n+7} ;] which i checked on Wolfram Alpha and symbolab and both say it's correct.

Now I have to plug it into [\; \lim ;] if I understand correctly. So: [\; \lim_{n->\infty} ;] [\; {(2n+5)(n+5)e}\over{2n+7} ;] = [\; \lim_{n->\infty} ;] [\; {2n}\over{2e} ;]

I know this is perfectly valid solution. But i have 3 similar examples which all end up going to [\; \infty ;] and that's why I have doubts if im doing it right.

2017.09.27 13:33 *Lookinfortips* **derivative math problem**

Using the definition of the derivatives to represent f(x) and a.

The limit is represents a derivative f'(a) lim ((4/(7+h) - 4/7)/h) h->0

I think the answer is, am i wrong? f(x)= 1/x a=1/7

If this doesn't right because of the format of reddit, I put it in a calculator, which I know wont help, but so it is easier to view the problem click the link. https://www.symbolab.com/solvelimit-calculato%5Clim_%7Bh%5Cto%200%7D%5Cleft(%5Cfrac%7B%5Cfrac%7B4%7D%7B7%2Bh%7D-%5Cfrac%7B4%7D%7B7%7D%7D%7Bh%7D%5Cright)

submitted by Lookinfortips to MathHelp [link] [comments]
The limit is represents a derivative f'(a) lim ((4/(7+h) - 4/7)/h) h->0

I think the answer is, am i wrong? f(x)= 1/x a=1/7

If this doesn't right because of the format of reddit, I put it in a calculator, which I know wont help, but so it is easier to view the problem click the link. https://www.symbolab.com/solvelimit-calculato%5Clim_%7Bh%5Cto%200%7D%5Cleft(%5Cfrac%7B%5Cfrac%7B4%7D%7B7%2Bh%7D-%5Cfrac%7B4%7D%7B7%7D%7D%7Bh%7D%5Cright)

2017.03.08 06:21 *Official_Not_Steve* **[Calculus] Limit and exponent problem**

I got a math problem on my homework

lim x->3+ (2(x-3))^{x-3}

Now in my mind I would just plug in 3 and have 0^{0} which turns into one but that seemed too easy. I then plugged it into the symbolab limit calculator which shows steps and it has a a much more complicated set of actions but also got one. Am I missing something or is my method correct?

submitted by Official_Not_Steve to MathHelp [link] [comments]
lim x->3+ (2(x-3))

Now in my mind I would just plug in 3 and have 0

2017.03.08 06:11 *Official_Not_Steve* **[Calculus] limits and exponents problem**

I got a math problem on my homework

lim x->3+ (2(x-3))^{x-3}

Now in my mind I would just plug in 3 and have 0^{0} which turns into one but that seemed too easy. I then plugged it into the symbolab limit calculator which shows steps and it has a a much more complicated set of actions but also got 1. Am I missing something or is my method correct?

submitted by Official_Not_Steve to askmath [link] [comments]
lim x->3+ (2(x-3))

Now in my mind I would just plug in 3 and have 0

2017.02.17 09:35 *TimeOfRa* **[College] Series involving cosine and D'Alembert's criterion**

Hello!

Here is the question:

We should test convergence of this serie

[; \sum_{n=2}^{\infty}\frac{cos\frac{\Pi *n^2}{n+1}}{ln^2n} ;]

We are supposed to do it with D'Alembert's criterion or Cauchy's criterion or Raabe

What I do is use D'Alembert's criterion ( and I get something lke this

[; \lim_{n\rightarrow \infty } \frac{cos\frac{\Pi (n+1)^2}{n+2}*ln^2n}{ln^2(n+1)*cos(\frac{\Pi n^2}{n+1})} ;]

And I don't what do to next, I don't know how to calculate limit of this. And I went to symbolab and they used [; -1\leq cos(\frac{\Pi n^2}{n+1})\leq 1 ;] and also integrals but we cannot use integrals for this. Why less/bigger than one? Should I do it like that too?

submitted by TimeOfRa to learnmath [link] [comments]
Here is the question:

We should test convergence of this serie

[; \sum_{n=2}^{\infty}\frac{cos\frac{\Pi *n^2}{n+1}}{ln^2n} ;]

We are supposed to do it with D'Alembert's criterion or Cauchy's criterion or Raabe

What I do is use D'Alembert's criterion ( and I get something lke this

[; \lim_{n\rightarrow \infty } \frac{cos\frac{\Pi (n+1)^2}{n+2}*ln^2n}{ln^2(n+1)*cos(\frac{\Pi n^2}{n+1})} ;]

And I don't what do to next, I don't know how to calculate limit of this. And I went to symbolab and they used [; -1\leq cos(\frac{\Pi n^2}{n+1})\leq 1 ;] and also integrals but we cannot use integrals for this. Why less/bigger than one? Should I do it like that too?