For $c &gt; 0,$ find the limit, lim$_{n \to \infty}  n(\sqrt[n]{c} - 1)$   Ok, I am not exactly on sure how to do this. Though here are some of my thoughts:    lim$_{n \to \infty} (nc^{1/n} - n) = nc^0 - n = 0,$ and I don't think this can be correct logically. Also, I think we can make it a $\log$ function by doing it like this:  $(nc^{1/n} - n) = e^{\log(nc^{1/n} - n)}.$ But, then what? We can't distribute the log inside because that's not correct. I am lost. Any help?   Update: please don't mark this as a duplicate of , since the other question was not required to use L'hopital's rule, but I need to use that in my question.
74526
How to find the limit of: $$ \lim_{x\to \infty }x\left(\sqrt[x]{a}-1\right)$$ Without using L'hôspital rule.  I've tried to bound the term and use the squeeze theorem but I couldn't find the right upper bound. I've also tried to convert $a^\frac{1}{x}$ to $e^{\frac{1}{x}\ln{a}}$ but it didn't help me.  Whats the right way to evaluate that limit?
65243
Calculate the limit: $$\lim_{x\to\infty} \sqrt[x]{3^x+7^x}$$  I'm pretty much clueless on how to approach this. I've tried using the identity of $c^x = e^{x \cdot \ln(c)}$ but that led me to nothing. Also I've tried replacing $x$ with $t=\frac{1}{x}$ such that I would end up with $\lim_{t\to 0} (3^{1/t} + 7^{1/t})^{1/t}$ however I've reached yet again a dead end.  Any suggestions or even hints on what should I do next?
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.
40298
In my homework I have to find limit (by interpreting it as an appropriately chosen function as the limit of integral sums):  $\lim_\limits{n\to \infty} \frac{1}{n}\sqrt[n]{(n+1)(n+2)(n+3)...(n+n)} (1)$. $$ $$ Solution:   $\lim_\limits{n\to \infty} \frac{1}{n}\sqrt[n]{(n+1)(n+2)(n+3)...(n+n)} = \lim_\limits{n\to \infty} ln(\frac{1}{n}\sqrt[n]{(n+1)(n+2)(n+3)...(n+n)}) = \lim_\limits{n\to \infty} ln(\frac{1}{n}\sqrt[n]{n^n(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})...(1+\frac{n}{n})}) = \lim_\limits{n\to \infty} ln(\frac{1}{n} n\sqrt[n]{(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})...(1+\frac{n}{n})}) = \lim_\limits{n\to \infty} ln((1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})...(1+\frac{n}{n}))^{1/n} = \lim_\limits{n\to \infty} \frac{1}{n} ln((1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})...(1+\frac{n}{n})) = \lim_\limits{n\to \infty} \frac{1}{n} \sum_{k=1}^{n} ln(1+\frac{k}{n}) = \lim_\limits{n\to \infty}  \sum_{k=1}^{n} \frac{1}{n} ln(1+\frac{k}{n})$ $$ $$ Note that function $f(x)=ln(1+x)$ is continuous in $[-1,1]$. Find integral $\int ln(1+x) dx = (x+1) ln(1+x)-x+C$.  Then $\int_{-1}^{1} ln(1+x)dx = ln(4)-2$. $$ $$ Assume that that limit $(1)$ is A. We show that $\forall \epsilon&gt;0$ $\exists N\in \mathbb{N}$: $\forall n\in \mathbb{N}$ $n\geq N \Longrightarrow $ $|\sum_{k=1}^{n} \frac{1}{n} ln(1+\frac{k}{n})-A|&lt;\epsilon$.  Suppose that $\mathfrak{T}$ is $[-1,1]$ set of subdivisions. Fix. $\epsilon &gt; 0$. That $f$ is integrative $[-1,1]$, then exist $\delta &gt;0$:   $\forall$ $T = T[x_0,...,x_n]\in \mathfrak{T}$ $\forall \xi=(\xi_1,...,\xi_n)\in[x_0,x_1]\times...\times[x_{n-1},x_n]$  $\lambda(T)&lt;\delta \Longrightarrow |\sum_{k=1}^{n} ln(1+\xi_k) (x_k-x_{k-1})-A|&lt;\epsilon$.  Which $N$ to choose? Is this right solution so far?
18978
Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$  Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?  $\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} &lt; \epsilon$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} &gt; \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \sqrt{n} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow n &gt; \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
110601
$$\lim\limits_{x\to 1}\frac{\sqrt x-x^2 }{1-\sqrt x}$$  I'm not entirely sure how to go about this, since rationalizing anything doesn't quite work for me. Just looking for some help.
24800
I am stuck on how to calculate the following limit: $$\lim_{x\to\infty}x\left(\left(1 + \frac{1}{x}\right)^x - e\right).$$  Definitely, it has to be through l'Hôpital's rule. We know that $\lim_{x\to\infty} (1+(1/x))^x = e$. So, I wrote the above expression as $$\lim_{x\to\infty}\frac{\left(1 + \frac{1}{x}\right)^x - e}{1/x}.$$Both numerator and denominator tend to zero as x tends to infinity. I applied l'Hôpital's rule twice, but I got the limit equal to infinity which is clearly wrong. In the book, it says that the limit should be $-e/2$.  Any help please? Also, in case someone managed to solve it, please tell me which numerator and denominator did you apply your l'Hôpital's rule to in both cases? Thanks a lot  Also, guys can you tell me how to write equations in this forum? Thanks.
99286
The limit is    $$\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$$  The problem is I don't know if I can calculate it normally like with a change of variables or not. Keep in mind that I'm not allowed to use L'Hôpital's rule nor the $\mathcal O$-notation.
58267
How to compute this limit:  $$\lim_{n\to\infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\cdots+\frac{n}{n^2+n^2}\right)$$  Please give me some hint.
77486
How can I find the following limit :$$\lim_{n\to\infty}\left(1-\frac{1}{3}\right)^2\left(1-\frac{1}{6}\right)^2\left(1-\frac{1}{10}\right)^2\cdots\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2$$There are four options $(a)\frac1{3}(b)\frac1{9}(c)\frac1{81}(d)0$ Thanks!
6393
I need to find simple limit: $$\lim_{n\to\infty}{\sqrt{n+1}-\sqrt{n}}$$  I can't use L'Hôpital's rule, which I think is natural for such problems. How to calculate such limit using only basic methods (definition and simple transformations etc.)?  It's one problem (from 50 problems) I don't know how to solve from my homework.  Thanks for help.
31748
How to find the value of $\lim_{n\to\infty}S(n)$, where $S(n)$ is given by $$S(n)=\displaystyle\sum_{k=1}^{n} \dfrac{k}{n^2+k^2}$$   is unable to calculate it.  This is a question from a questions booklet, and the options for the answer are--  $\begin{align} &amp;A) \dfrac{\pi}{2} \\ &amp;B) \log 2 \\ &amp;C) \dfrac{\pi}{4} \\ &amp;D) \dfrac{1}{2} \log 2 \end{align}$
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.
64033
I am trying to calculate the limit of  $$\lim_{n \to \infty} \frac{(2n-1)!}{(2n+1)!}$$  and I am stuck. I tried n division but that didn't work. Any help is appreciated
102365
Find the following limit: $$ \lim_{x\to \infty} (\sqrt{x^2+x}-\sqrt{x^2-x} )$$   I tried to simplify using conjugation. This gave me the following: $$ \lim_{x\to \infty} \frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}} $$  When I plug in the $\infty$, I'm left with $ \frac{\infty}{\infty} $. Did I mess up somewhere, or does the limit not exist?
66365
I am facing difficulty with the following limit. $$ \lim_{n\to\infty}\left(\binom{n}{0}\binom{n}{1}\dots\binom{n}{n}\right)^{\frac{1}{n(n+1)}} $$ I tried to take log both sides but I could not simplify the resulting expression. Please help in this regard. Thanks.
69669
For $a,b,c&gt;0$, Find $$ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$$  how can I find the limit of sequence above?  Provide me a hint or full solution. thanks ^^