Somebody asked this and I think it's quite interesting as I couldn't figure out how to evaluate this but the Wolfram Alpha says its limit is $\frac e2$. $$\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$$ Could someone help here?
7188
Find the following limit: $$\lim_{x\to 0}\frac{(1+x)^{1/x}-e}{x} \tag{1}$$  The following is my approach, although is full of incorrect assumptions (statements etc).  $$f(x)= \lim_{h\to 0}(1+x+h)^{1/(x+h)}\\$$  From here we can say $f(0) = e$.  $$\ln(f(x)) = \lim_{h\to 0}\frac{1}{x+h}\ln(1+x+h)$$  Near $x=0$, we can use the series of logarithm: $$\ln(f(x)) = \lim_{h\to 0}\frac{1}{x+h}\left(x+h-\frac{(x+h)^2}{2}...\right)$$  Differentiating we get:  $$\frac{f'(x)}{f(x)} = \lim_{h\to 0}-\frac{1}{2} +\frac{x+h}{3}  ... \tag{2}$$  Now we note that $(1)$ is actually $f'(0)$ (?) and so from $(2)$ we get:  $$f'(0) =\frac{-1}{2} f(0) = \frac{-e}{2}$$  While the answer is seemingly correct, the method is absolutely not. It looks like this is fluke than anything else.  I also tried computing taylor series of $(1+x)^{1/x}$ near $x=0$ but couldn't do it.
60294
How to solve the following limit question? $$\lim_{n\to\infty}\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}$$  Thanks a lot.
9394
Find the following limit  $$ \lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2} $$  I have used natural logarithm to get  $$ \exp\lim_{x\to0}\frac1{x^2}\ln\left(\frac{1+x2^x}{1+x3^x}\right) $$  After this, I have tried l'opital's rule but I was unable to get it to a simplified form.  How should I proceed from here? Any here is much appreciated!
5228
I'm supposed to calculate: $$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}$$ By using WolframAlpha, I might guess that the limit is $\frac{1}{2}$, which is a pretty interesting and nice result. I wonder in which ways we may approach it.
98641
How do I calculate $\lim_{x\to 1}\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}$.  Please help me.  Thanks!
15280
Can anyone help me with this limit:  $$ \lim_{n \to \infty}{\left(e^{\frac{1}{n}} - \frac{2}{n}\right)^n}$$  I think that I should expand $e^{\frac{1}{n}}$ as the sequence $1 + \frac{1}{n} + \frac{1}{n^{2}2!} + \frac{1}{n^{3}3!} + \dots $. But I can't see how it helps me.  Edit: or maybe use logarithm?
72097
$$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$   by directly substituting $x=0$ i got $\infty$ by using L-H's rule, i got $-1/8$   the given options are  $a)\frac{24e}{11}$    $b)\frac{11e}{24}$    $c)\frac{e}{11}$    $d)\frac{e}{24}$    may be the question would be  $$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$
70007
now I am evaluating limits of functions, but i dont know how to start to solve this limit. It is possible without L Hopital's rule?  $\lim_{x\to0}{{\frac{\ln(\cos(4x))}{\ln(\cos(3x))}}}$?
17931
On Wikipedia, it says that $\lim_{n\to\infty}(1+\frac{1}{n})^n = e$ :     It [e] is approximately equal to 2.71828,[1] and is the limit of (1 + 1/n)n as n approaches infinity, ... ()   When I evaluate $(1+\frac{1}{n})^n$ for $n = 10^8$, I get approximately $2.718281798347$ which indeed is pretty close to $e$.  But when I try to "solve" the limit using the laws of limits, I get  $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = \left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)\right)^n$$  because of the power law  $$ = \left(\lim_{n\to\infty}(1)+\lim_{n\to\infty}\left(\frac{1}{n}\right)\right)^n$$  because of the addition law  $$=\left(1+0\right)^n = 1$$  but that would mean that $e=1$, which is obviously not true.  What am I missing / doing wrong?  Thanks in advance.
62679
$$\lim_{x\to4} \frac{1}{x-4}$$ Would it be correct to say that the limit is undefined because the denominator would be $0$?
59163
I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Is it possible to compute this limit with the McLaurin expansion? Can you explain the method and the steps used? Thanks. (I prefer to avoid to use L'Hospital's rule.)  $$\lim _{x\to 0}\frac{x\bigl(\sqrt{3e^x+e^{3x^2}}-2\bigr)}{4-(\cos x+1)^2}$$
32524
How can I find the $\lim  (1+(a^{1/n}-1) / b)^n$ when $n \to \infty$?    I would greatly appreciate it if you kindly give me some hints.
31010
I tried to evaluate $$\lim_{x\to0}\frac{(1+x)^{1/x}-e}x$$but all my efforts were in vain . L'Hospital rule becomes very complicated.
57957
Purely out of curiosity, is it possible to compute the following limit without using L'Hopital?  $$\lim_{x\to 0} \frac{e^{-1/x^2}}{x}$$  Using L'Hopital and realizing that we cannot get what we want if we simply take derivatives of $e^{-1/x^2}$ and $x$ directly, we have that  $$\lim_{x\to 0}\frac{e^{-1/x^2}}{x}=\lim_{x\to 0}\frac{1}{x}\frac{1}{e^{1/x^2}} = \lim_{x\to 0} \frac{1/x}{e^{1/x^2}}=0$$  I am curious about whether there is a different way of computing this.
73977
I have to find out the following limit $$\lim_{m\to\infty}\lim_{n\to\infty}[\cos(n!πx)^{2m}]$$ for $x$ rational and irrational. for $x$ rational $x$ can be written as $\frac{p}{q}$ and as $n!$ will have $q$ as its factor the limit should be equal to 1. the second part of irrational is giving me problems. I first thought that limit should be zero as absolute value of cosine term is less than 1 and power it to infinity you should get $0$. But then I realised that it was wrong. I brought the limit down to this form. $$e^{-\sin^2(n!πx)m}$$ after this I find the question quite ambiguous as they have just said $x$ is irrational. If I take $x$ as $\frac{1}{n!π\sqrt{m}}$ I get the limit as $\frac{1}{e}$ but if I take  $x$ as $\frac{2}{n!π\sqrt{m}}$ I get the limit as $\frac{1}{e^4}$. please help me and tell me where have I gone wrong?
118596
This was the question of a test. My question is if my attempt to solve it is correct, and if it is, why is it correct.  $$\lim_{(x,y) \to (0,0), x+y \neq 0}{\frac{\ln(1-x-y)}{x+y} } $$  My attempt:   Let $\xi = -x-y $. Then $\xi \to 0$ whenever $(x,y) \to (0,0)$ and $x+y \neq 0 \iff \xi \neq 0$. (Is it then correct to say that the previous limit exists and is equal to the following iff the following exists? And why?):  $$\lim_{\xi \to 0, \xi \neq 0}{\frac{\ln(1+\xi )}{-\xi}}$$   If it is correct, then the limit exists and is $-1$. If it is correct, why is it correct?
86721
Calculate $$\lim_{x\to 0}\dfrac{1}{x^{\sin(x)}}$$  I'm pretty much clueless here, only that there is L'hospital obviously here.   Would appreciate any help.