I'm looking for the answer to this question. But I could not find the "satisfactory" answer.  This is obvious,     $$\lim_{n\to \infty} \left(1+\frac 1n \right)=1+0=1$$   and     $$\lim_{n\to \infty} \left(1+ \frac 1n \right)^n=e≈2.718281...$$   And also we can write,     $$\lim_{n\to \infty} \left(1+ \frac 1n \right)^n=\left(1+0 \right)^{\infty}=1^{\infty}≠1$$   The point, which that I can not understand is:     $$\lim_{n\to \infty} 1^{\infty}=1^{\infty}=1$$   But, this is also true:     $$\lim_{n\to \infty} \left (1+\frac 1n \right) =\lim_{n\to\infty}1=1$$      Why $\lim_{n\to \infty} \left(1+ \frac 1n \right)^n=e$ doesn't imply $\lim_{n\to \infty} \left(1+\frac 1n \right)≠1?$
83864
In fact the real question is: Does $\lim\limits_{n\to\infty}1^{n}=e$?.  I know that   $$ \lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n=e, $$  So, can we say that $1^\infty=e$?  And, by logic, This product  $$\underset{\text{infinity times}}{1\cdot1\cdot\ldots\cdot1},$$ gives $1$ not other value.
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.
84235
I'm asked to prove that $$\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\frac{\pi}{4}$$ This looks like it can be solved with Riemann sums, so I proceed:  \begin{align*} \lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)&amp;=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{n}{n^2+k^2}\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{n^2}{n^2+k^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{1}{1+(k/n)^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}f(\frac{k}{n})(\frac{k-(k-1)}{n})\\ &amp;=\int_{0}^{1}\frac{1}{1+x^2}dx=\frac{\pi}{4} \end{align*}  where $f(x)=\frac{1}{1+x^2}$. Is this correct, are there any steps where I am not clear?
120017
$$\lim_{n \to \infty} \frac{n!}{n^{n}}$$    Attempt:  $$n!=n\left( n-1 \right)\left( n-2 \right)...\left( n-\left( n-1 \right) \right)=n^{n}+...$$  $$\lim_{n \to \infty} \frac{n!}{n^{n}}=\lim_{n \to \infty} \frac{n^{n}+...}{n^{n}}=\lim_{n \to \infty} \frac{1+...}{1}=1+\lim_{n \to \infty} \ ...=1$$    I based this on the fact that $n$ is the highest power. The answer's meant to be $0$ though.
42353
I’m having a problem approaching the question. The fact that it’s a limit of a sum is confusing to me.   $$\lim _{n\to \infty }\left(\frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}\right)$$
66782
I'm trying to find $$ \lim_{n \to \infty}   \int_1^a \dfrac{n}{1+x^n} \, dx  ,$$ where parameter $a&gt;1$. The limit appears to be $\ln 2$ regardless of the value of $a$. Some trick is probably needed here, but i don't see it.
78897
$\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$  I let, $x = \frac{1}{n}$, then as   $\lim_{x \to 0} \frac{1}{x}[(1+x)^\frac{1}{x} - e] = \infty$  L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{1} = \frac{1}{0}(1+0)^{\frac{1}{0}-1} = \infty$  Again, if we apply L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{x+(1+x)}$. This is also going to $\infty$ as $x \to 0$. But, I know the answer is $\frac{-e}{2}$, and I am not even close. Can anyone please find me the mistakes here. Oh, I am supposed to use L'Hopital's rule.
114120
How can I calculate this: $\lim\limits_{n \rightarrow \infty} \frac{1+\sqrt[n] {e}+\sqrt[n] {e^2}+ \dots + \sqrt[n] {e^{n-1}}  } {n}$? I how come so far: $$ \lim_{n \rightarrow \infty} \frac{1+\sqrt[n] {e}+\sqrt[n] {e^2}+ \dots + \sqrt[n] {e^{n-1}}  } {n} = \lim_{n \rightarrow \infty} \frac{1}{n} \left(1 +\sqrt[n] {e}+\sqrt[n] {e^2}+ \dots + \sqrt[n] {e^{n-1}}\right) = $$  $$ \lim_{n \rightarrow \infty}\frac{1}{n}\sum_{i =1}^{n}e^{\frac{i-1}{n}}$$ But how do I continue?
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  $$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$
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?   $$\lim_{n\to\infty} \left[ n-{n\over e}\left(1+{1\over n}\right)^n \right].$$
70232
I'm asked to find $\displaystyle\lim_{n\to \infty}\sum_{k=1}^n \left(\frac{k}{n}\right)^n$.  It seems that the sum converges.  I managed to prove that $\forall n,1&lt;\sum_{k=1}^n \left(\frac{k}{n}\right)^n&lt;e$  But I don't know how to prove convergence, nor find the value of the limit.
23690
Since $\infty&gt;0$, so $1/\infty &gt;0$, thus I think $1/\infty$ should be infinitesimal, but the calculus book says  $$\lim_{x \to \infty} \frac{1}{x}= 0$$  So is $1/\infty$ zero or infinitesimal ?  P.S. I mean are $1/\infty$ and $\lim_{x \to \infty} 1/x$ the same thing here?
17446
I've seen some limit problems where you can do this: $$ \lim_{x \to \infty} \exp\left({g(x)}\right) = \exp \left( \lim_{x \to \infty }g(x)\right) . $$ So, I've tried to generalize the result as: $$ \lim_{x \to \infty} f(g(x)) = f \left( \lim_{x \to \infty }g(x) \right) $$ and I was just wondering if the result above is actually true, and on what conditions? Perhaps someone can point me to a theorem.
23944
I can't get this to $ = \frac{0}{0}$ form so I can use l'Hospital rule $$\lim_{x \to \infty} x - x^{2}\ln\left(1 + \frac{1}{x}\right)$$  tips?  [EDIT] $$\lim_{x \to 0} \frac{1}{x} - \frac{\ln(1 + x)}{x^{2}}$$ second term $$\lim_{x \to 0} \frac{\ln(1 + x)}{x^{2}} =  \lim_{x \to 0}  \frac{\frac{1}{1 + x}}{2x} = 2\lim_{x \to 0}  \frac{x}{x+1} = 0$$ first term $$ \lim_{x \to 0}  \frac{1}{x} = \infty$$  is this okey?
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?
6398
How does one show that $\lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right)^x$ converges? I have a book that uses this to define $e$ but they don't show that it converges. I suppose we could give it an upper and lower bound but I don't know how to do this. I've seen this question before on here but I haven't seen an answer I like yet.
49038
I know that $\lim_{x\to0}\frac{x}{x}=$ 1.  But in my text book, it is written that it is $\infty$ and even $\frac10=\infty$.  But how is it possible? And are they both same? What is the difference between indeterminate and undefined?  And if they are $\infty$ then what is the reason? And why can't we define something for these numbers?    Now see that book says when $x=0$, $t=\infty$. If it is wrong, why did they just write it?