I am trying to evaluate $$\lim\limits_{x\to \infty}x^2 \left( a^{1 \over x} - a^{1 \over x+1} \right)$$ for $a&gt;0$.  The idea: $$a^{1 \over x} = e^{\ln{a^{1 \over x}}}=e^{{1 \over x}\ln{a}} = 1 + {\ln{a} \over x} + {1 \over 2}{(\ln{a})^2 \over x^2}+{1 \over 6}{(\ln{a})^3 \over x^3}+\dots$$  $$a^{1 \over x}-a^{1 \over x+1} = e^{\ln{a^{1 \over x}}}-e^{\ln{a^{1 \over x+1}}}=e^{{1 \over x}\ln{a}}-e^{{1 \over x+1}\ln{a}} = \ln{a} \left({1\over x} - {1 \over x+1} \right)+ {1 \over 2}(\ln{a})^2\left({ {1\over x^2} - {1 \over (x+1)^2}}\right)+{1 \over 6}(\ln{a})^3\left({ {1\over x^3} - {1 \over (x+1)^3}}\right)+\dots$$  $$x^2 \left(a^{1 \over x}-a^{1 \over x+1}\right) =\ln{a} \left(x - {x^2 \over x+1} \right)+ {1 \over 2}(\ln{a})^2\left({ 1 - {x^2 \over (x+1)^2}}\right)+{1 \over 6}(\ln{a})^3\left({ {1\over x} - {x^2 \over (x+1)^3}}\right)+\dots$$  This is where I get stuck. I suppose the whole sum should converge to $\ln{a}$ but can't quite find the way to get there. I would appreciate any comments.
123790
I have a little problem with limit of this function:  $\lim_{x \to \infty} x^2(2017^{\frac{1}{x}} - 2017^{\frac{1}{x+1}})$  I have tried de L'Hopital rule twice, but it doesn't work. Now I have no idea how to do it.
23242
Let $a_1, \ldots, a_n &gt; 0$. How to compute the limit $\lim_{n \to \infty} \sqrt[n]{a_1^n + \cdots + a_m^n}$?  My solution:  $$ \lim_{n \to \infty} \sqrt[n]{a_1^n + \cdots + a_m^n} \\ = \lim_{n \to \infty} (1 + a_1^n + \cdots + a_m^n - 1)^{1/n} \\ = \lim_{n \to \infty} \left( (1 + a_1^n + \cdots + a_m^n - 1)^{\frac{1}{a_1^n + \cdots + a_m^n - 1}} \right)^{ \frac{a_1^n + \cdots a_m^n-1}{n}}. $$ But then we need to have the condition $a_1^n + \cdots + a_m^n - 1 \to 0$ ($n \to \infty$).   Thank you very much.
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?
24422
This is being asked in an effort to cut down on duplicates, see here: , and here: .    What methods can be used to evaluate the limit $$\lim_{x\rightarrow\infty} \sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x.$$  In other words, if I am given a polynomial $P(x)=x^n + a_{n-1}x^{n-1} +\cdots +a_1 x+ a_0$, how would I find $$\lim_{x\rightarrow\infty} P(x)^{1/n}-x.$$  For example, how would I evaluate limits such as $$\lim_{x\rightarrow\infty} \sqrt{x^2 +x+1}-x$$ or $$\lim_{x\rightarrow\infty} \sqrt[5]{x^5 +x^3 +99x+101}-x.$$
34311
How can I show:      $$\lim_{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$$   I've tried multiplying with its "conjugate" but that doesn't seem to help that much.  Thanks!
83831
I dont get it, if I look at $$\lim \limits_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$$    It seems like it should be $1$ because $1 \over n$ is almost zero and $1^\infty$ is still $1$
97153
I want to evaluate $$\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right).$$  I have computed  \begin{align*} a_{n+1}-a_n &amp; = \frac{1}{\sqrt{(2n+1)(2n+2)}}-\frac{1}{\sqrt{n(n+1)}}\\             &amp; = \frac{1}{\sqrt{n+1}}\left( \frac{1}{\sqrt{2(2n+1)}}-\frac{1}{\sqrt{n}} \right). \end{align*} Now I'm stuck.
41326
I am to find the limit of  $$\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$$  so I used: $$\lim_{x \to -\infty} = \lim_{x \to \infty}f(-x)$$  but I just can't solve it to the end.
117031
$$\lim\limits_{x\to\infty} \ \frac{\ln \ x}{x}  = 0$$  How can I know this without looking at the graph? What's the easiest way to equate that expression to 0?  I understand that an integer over x approaches 0 as x approaches 0, but since ln x is an increasing function as x approaches infinity, how do we know which effect outweighs the other?
5779
I want to calculate the limit of: $$\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $$ or prove that it does not exist. Now I know the result is $4$, but I am having trouble getting to it. Any ideas would be greatly appreciated.
11378
I have managed to use the root test and get as far as getting the limit to equal to $1$. But, I still need clarification on why the answer for the limit to this questions is $\alpha &lt; 1$. \begin{align} \sum _{n=0}^{\infty }\left( \frac {\alpha n} {n+1}\right) ^{n}\end{align} \begin{align} \lim _{n\rightarrow \infty }\left( \left( \frac {\alpha n} {n+1}\right) ^{n}\right) ^{1/n} =\left| \frac {\alpha n} {n+1}\right| =\alpha \lim _{n\rightarrow \infty }\left| \frac {n} {n+1}\right|=1.\end{align}
88599
Find   $$\lim _{x\to \infty }x\Big(\ln\left(x+1\right)-\ln x\Big)$$  Here's how I do it:  $$x\Big(\ln(x+1)-\ln x\Big) = x\Bigg(\ln(x(1+\frac{1}{x}) - \ln x\Bigg)$$ $$x\Big(\ln x + \ln(1+\frac{1}{x}\Big)-\ln x =x\ln\Big(1+\frac{1}{x}\Big)$$ $$\ln\Bigg( \Big(1+\frac{1}{x}\Big)^x \Bigg) \rightarrow \ln 1^\infty = 0$$  What am I doing wrong? The answer is supposed to be $1$, but I get $0$.
64999
This was a practice problem I was given.      Show that $\lim_{n \rightarrow \infty} \sqrt[n]{a} =1$.   My argument is as follows: \begin{align} \lim_{n \rightarrow \infty} \sqrt[n]{a} &amp;= \lim_{n \rightarrow \infty} e^{\ln(\sqrt[n]{a})} \\ &amp;= e^{\lim_{n \rightarrow \infty}(\frac{1}{n}) \ln(a)} \\ &amp;= e^{\lim_{n \rightarrow \infty}(0 \cdot \ln(a))} \\ &amp;= e^0 \\ &amp;= 1.\end{align}
29010
Given $\lim \limits_{x\to\infty}(1+\frac{1}{x})^{x}$, why can't you reduce it to $\lim \limits_{x\to\infty}(1+0)^{x}$, making the result "$1$"?  Obviously, it's wrong, as the true value is $e$.  Is it because the $\frac{1}{x}$ is still something even though it's really small?  Then why is $$\lim_{x\to\infty}\left(\frac{1}{x}\right) = 0\text{?}$$  What is the proper way of calculating the limit in this case?
122981
Suppose $a \in \mathbb{R} ,a&gt;1$ Is there an idea to compute the limit below ? $$\lim_{n \to \infty}\left( \frac{1^a+2^a+3^a+...+n^a}{n^a}-\frac{n}{a+1} \right)$$ I tried it for $a=1,2,3$ but I get stuck in general form . Can someone help me ? Thanks in advance.
7014
Hello I was hoping that someone might be able to verify that the following proof that $\lim_{x\to 0} {1\over x}$ does not exist is correct.   First assume that $\lim_{x\to 0} {1\over x} = L$. This means that for every $\epsilon &gt; 0$ there is a $\delta &gt; 0$ such that for all $x$ if $0 &lt; \lvert x\rvert &lt; \delta$ then $\lvert{1\over x} - L \rvert &lt;\epsilon.$  Now let $\epsilon=1$ and choose $x &lt; \min(1,\delta)$.  Therefore there is a $\delta&gt;0$ such that if $0 &lt; \lvert x\rvert &lt; \delta$ then $\lvert{1\over x} - L \rvert &lt;\epsilon$.  It follows that $\lvert{1\over x}\rvert &lt; 1+ \lvert L\rvert$.  However clearly $\lvert{1\over x}\rvert&gt;1$ and therefore a contradiction has been reached and there is no number $L$ such that $\lim_{x\to 0} {1\over x} = L$.
19573
Let $a &gt; 0$ and let $k \in \mathbb{N}$. Evaluate: $$\lim_{n\to \infty}a^{-nk}\prod ^k_{j=1}\left(a+\frac{j}{n}\right)^n$$  Clueless on this problem. Seek your help.