Taken from :     The number $e$ is the limit $$e = \lim_{n \to \infty} \left (1 + \frac{1}{n} \right)^n$$   Graph of $f(x) = \left (1 + \dfrac{1}{x} \right)^x$     Its evident from the graph that the limit actually approaches $e$ as $x$ approaches $\infty$. So I tried approaching the value algebraically. My attempt:  $$\lim_{n \to \infty} \left (1 + \frac{1}{n} \right)^n$$ $$= \lim_{n \to \infty} \left(\frac{n + 1}{n}\right)^n$$ $$= \left(\lim_{n \to \infty} \left(\frac{n + 1}{n}\right) \right)^n$$ $$= 1^\infty$$  which is an indeterminate form. I cannot think of any other algebraic manipulation. My question is that how can I solve this limit algebraically?
48461
How to prove the following limit?   $$\lim_{n \to \infty} (1+1/n)^n = e$$    I can only observe that the limit should be a very large number!   Thanks.
113156
The limit I have to evaluate is this -  $$\lim_{x \to \infty} \frac{e^x}{\left(1+\frac{1}{x}\right)^{x^2}}$$  I first checked if L'Hopital's rule applies here. The limit of both numerator and denominator is $\infty$. But differentiating the denominator yield a even more complicated expression.  I am not getting how to approach this question using some other method. Thank you.
109013
How would I work out a limit of the form:  $$\lim_{x\to 0}\;(1+x)^{1/x}$$  I know these types of limits have a solution based on $e$ but how do I find this solution?
47565
I get that the limit is  $$\lim\limits_{n\to\infty}f(x)=1$$ because $$\lim\limits_{n\to\infty}(1+x^{2n})=1,\,\,\lvert x\rvert&lt;1.$$  Is this right?
95669
Many websites and calculus books give this well known result \begin{equation} \lim\limits_{n\to\infty} \left(1 + \frac{x}{n}\right)^n = e^x \end{equation} However, a textbook I was reading casually mentioned that if $x_n \rightarrow x$ then  \begin{equation} \lim\limits_{n\to\infty} \left(1 + \frac{x_n}{n}\right)^n = e^x \end{equation} Why is this true? It seems very intuitive but I feel some explanation is missing. Thank you!
72860
I'm working on a practice exam for my masters quals and I am having difficulties with the following limit. According to wolfram alpha, it's value is 1/2. Does anyone know how to find this limit?   $$\lim_{n\to\infty} \left[ n-{n\over e}\left(1+{1\over n}\right)^n \right].$$
117850
When calculating the limit $L=\big( 1 - \frac 1 {\sqrt n}\big)^n$, $n\to\infty$, what allows me to do the following:  $$ L=\lim \left(\left(1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n} $$  As the term inside the outer parenthesis goes to $e^{-1}$, we have $L=\lim e^{-\sqrt n}=0$.  It's like we're distributing the parenthesis somehow:  $$\lim \left(\left(1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}=\lim \left(\lim\left (1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}=\lim e^{-\sqrt n}$$  The question is: Why can we do this? Which property are we using?
109362
Evaluate the limit      $$ \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} $$   My Attempt:  $$ \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} = \lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^{n^2}\cdot \frac{1}{e^n} $$  How can I solve the problem from this point?
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.
31094
$$\lim_{n\to \infty}2^{\frac{1}{n}}$$   Basically looking at $$\lim_{n\to \infty}2^{\frac{1}{n}}=2^{\lim_{n\to \infty}{\frac{1}{n}}}=2^0=1$$  But if I use $$\lim_{n\to \infty}2^{\frac{1}{n}}=\lim_{n\to \infty}e^{\ln\left(2^{\frac{1}{n}}\right)}=\lim_{n\to \infty}e^{{\frac{1}{n}}\cdot{\ln(2)}}=\lim_{n\to \infty}e^{\frac{1}{n}}\cdot\lim_{n\to \infty}e^{\ln(2)}=1\cdot2=2$$  Where did I got it wrong?
47849
Is there a way to find this limit without using L'Hôpital's rule. Just by using some basic limit properties.  $$\lim_{x\to\infty}\left(1+\frac1x\right)^x=e$$
124045
Show that, $$e=\lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x $$  Is the same number that satisfies,  $$\lim_{h\to0} \frac{e^{h}-1}{h} = 1 \tag{*}$$  You don't have to do that if it's too cumbersome, but (*) is used to find the derivatives of real exponential functions. I know that definitions are not to be proven but I'm looking for some derivation or intution for the claim that (*) is a legitimate definition for e, and furthermore, a proof that limit in (*) exists.
41452
The value of $$\lim_{n\to \infty} \int_0^1 nx^n e^{x^2} dx$$ is ____________(round off to three decimal places) I tried integrating by parts and bring out some recurrence relation , but it was of no use. However, I can show the limit to be greater than or equal to 1. $\int_0^1 nx^n e^{x^2} dx \gt \int_0^1 nx^{n+2} dx$ (using  $e^x\gt x $ for $x\gt 0$) $=\frac n{n+3}=1-\frac 3{n+3}$ Taking limit as $n\to \infty$ , proves my assertion. Please help me find the actual limit.Thanks in advance.
96736
Today in my analysis class, we were preparing for the final and this question came up:     Evaluate $$\lim_{x\to0}\frac{e^{-1/x^2}}{x}$$   We tried taking the $\log$, using L'Hopitals and some other tricks but couldn't figure it out.   I thought maybe viewing this limits as a sequence in the following way might help;  $$\lim_{x\rightarrow0}\frac{e^{-1/x^2}}{x} = \lim_{n\rightarrow \infty}\frac{e^{-n^2}}{1/n} = \lim_{n\rightarrow \infty}\frac{n}{e^{n^2}}$$  But from them I'm not sure.   Thank you.
23066
How to compute the following limit?  $\lim\limits_{n\to\infty} n&#183;(\sqrt[n]{a}-1)$  We know that it's got something to do with $\ln$ or $\exp$.  We know that $\lim\limits_{n\to\infty} \sqrt[n]{a} = 1$ but it seems to not to be true that therefore $\lim\limits_{n\to\infty} n&#183;(\sqrt[n]{a}-1) = \lim\limits_{n\to\infty} n&#183;(1-1) = 0$.  What we know is  $\lim\limits_{n\to\infty} (1-\frac{1}{n})^n = \lim\limits_{n\to\infty} (1 + \frac{1}{n})^n \lim\limits_{n\to\infty} (1-\frac{1}{n})^{n+1} = e$  and  $\lim\limits_{n\to\infty} (1+\frac{x}{n})^n = \exp(x)$
62311
Here's the problem... $$\lim\limits_{x\to \infty}\ (x!)^{1/x}$$  I've deduced the answer to this is $\infty$, but haven't exactly shown that.  I'm getting $\infty^0$, so did both tricks where you change it to  $$\exp\left({\lim_{x\to\infty}} \frac{1/x}{1/\ln(x!)}\right) = \exp\left(\frac{0}{0}\right)$$ And $$\exp\left({\lim_{x\to\infty}} \frac{\ln(x!)}{1/(1/x)}\right) = \exp\left(\frac{\infty}{\infty}\right)$$  And for both of which, I need to do L'Hospital on this $\ln(x!)$ which I haven't the slightest idea how to do... Any suggestions that don't include Stirling's Approximation? My prof eluded to this being out of the question unless we were going to prove Stirling's first
90860
We know that $\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n=e$ and so I thought the approach to evaluating the limit in the question would be to just use this fact and substitute it into the numerator. This approach would tell us the above limit evaluates to $1$. However, that does not seem to be the correct limiting value. In fact, it evaluates to $\frac {1}{\sqrt{e}}$. Why is this so?