Compute  $$\lim_{n \to \infty}n \int_0^{\frac{\pi}{4}}\tan^n x\,dx$$  I tried to define a recurrence with $I_n=\int_0^{\frac{\pi}{4}} \tan^n x$ :  $I_0 = \frac{\pi}{4}, I_1=\ln\sqrt{2}$  and $I_{n}=\frac{1}{n-1}-I_{n-2}$, but I can't complete it.
3624
$$f(x)=\lim_{n\to\infty}\biggl(n\int_0^{\pi/4}(\tan x)^n\,dx\biggr)$$ I try to this way, $\tan x\ge x$, when $x\in(0,\frac\pi4)$, but this turns out to be $\tan x\ge0$, which is obvious even without calculation. I think it can be solved by using the squeeze rule but I can't find a proper scaler to meet $A=g(x)\le f(x)\le h(x)=A$ when $n\to\infty$   $\color{red}{Edit}$: According to@Khallil's advice, I can solve that $f(n)+f(n+2)=\frac1{n+1}$, and it's trivial that $\frac1{n+1}=f(n)+f(n+2)\le 2f(n)\le f(n-2)+f(n)=\frac1{n-1}$. So $f(x)=\frac12$.  This is easy and quick. Btw, is there any other idea?  $\color{red}{Edit[2]}$: I find that @Byron Schmuland 's method is especially useful in a specific kind of problem. $$f(x)=\lim_{n\to\infty}\int_0^1\frac{x^n}{1+x}dx$$,for example, Let $X_,...,X_n$ be i.i.d. random variables with density$f(x)=1$ on $(0,1)$, CDF(cumulative density function) of $X$ is $F(x)=x$. Now let $M=\max(X_1,\dots,X_n)$, its density function is: $f_M(x)=nx^{n-1}$ for $x\in(0,1)$ Also, it's also not hard to see that $M\to1$ in distribution as $n\to\infty$. So,$$f(x)=\lim_{n\to\infty}\int_0^1\frac{x^n}{1+x}dx=\lim_{n\to\infty}\frac1n\int_0^1{f_M(x)\frac{x}{x+1}dx}=\lim_{n\to\infty}\frac1nE(\frac{M}{M+1})=0$$ Although this simple example can be solved by other ways more easily, but this gives us another perspective.
13847
I am trying to compute this integral/limit, I don't feel like I have any good insight...   $$\lim_{n \to \infty} \int_0^{n^2} e^{-x^2}n\sin\frac{x}{n} \, dx.$$  I have tried to make a change of variable to get rid of the $n^2$, I changed to $X=\frac{x}{n^2}$ but got something even worse, I've tried to reach a situation where I could use a convergence theorem for Lebesgue Integrals,... I'm not sure I'm even on the right track!  Could you give me a hint on how to start this?  Thank you very much!
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?
75155
So I came across this question on brilliant.org and didn't know how to go about it: $$\int_{0}^{\infty} \frac{ \ln(x)}{x^2+2x+4} dx $$  I tried to complete the squares in the denominator and then use a trigonometric substitution (with tan) but I didn't get anything beyond that.   How do I do it? Do we need to differentiate under the integral sign here?
69234
Find $\lim_{n\to \infty} \int_n^{n+1} {\sin x \over x}  dx$   I thought about defining $\space F(x) = \int_0^x {\sin t \over t}  dt \space$ and then the limit is $\space \lim_{n\to \infty} (F(n+1) - F(n))$, so the answer is 0? It doesn't seem ok
69767
$$\lim_{n\to\infty}1+\sqrt[2]{2+\sqrt[3]{3+\sqrt[4]{4+\ldots+\sqrt[n]{n}}}}$$  I have no idea about this.  The equation can be written in its recursive form as:  $$f(n) = g(1,n)$$  Where  $$g(x,n) = [x\impliedby n]\cdot (x+ g(x+1,n))^{\frac 1x}+[x=n]\cdot (n)^{\frac 1n}$$  Of course, [] is the indicator function representing of piece wise notation.
87113
As the title suggests, I am interested in computing $$\lim_{x \to \infty} \frac{\sin x}{\sin x}.$$ Some of my friends say that the limit is $1$ because the domain of function is $\mathbb{R} \setminus \{n \pi : n \text{ is an integer}\},$ whereas others say that this limit is undefined. Which opinion is correct?
4262
fine the limit :  $$\lim_{ n \to \infty }\frac{1}{n}\int_{0}^{n}{ \frac{x\ln(1+\frac{x}{n})}{1+x}}=?$$  My Try: in the  $$I=\int_{}^{}{ \frac{x\ln(1+\frac{x}{n})}{1+x}}=\left(x+n\right)\ln\left(\left|x+n\right|\right)+\left(\ln\left(n\right)-\ln\left(n-1\right)\right)\ln\left(\left|x+1\right|\right)+\operatorname{Li}_2\left(-\dfrac{x+1}{n-1}\right)+\left(-\ln\left(n\right)-1\right)+c$$  now?
107337
I just came across this limit and I suppose it can be computed using a Riemann sum but I can't get it right. $$\lim _{n\to \infty }\sum _{k=1}^n\sqrt{n^4+k}\sin\left(\frac{2k\pi }{n}\right)$$ Any ideas?
29585
$\lim\limits_{n\to\infty}{\sum\limits_{k=n}^{5n}{k-1 \choose n-1}(\frac{1}{5})^{n}(\frac{4}{5})^{k-n}}$  It's clear that we can simplify the limit a little bit, after which we get:  $\lim\limits_{n\to\infty}{(\frac{1}{4})^{n}\sum\limits_{k=n}^{5n}{k-1 \choose n-1}(\frac{4}{5})^{k}}$  I could further simplify the expression, but I feel like there's a more elegant solution.   Give me a hint, please
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--  $\begin{align} &amp;A) \dfrac{\pi}{2} \\ &amp;B) \log 2 \\ &amp;C) \dfrac{\pi}{4} \\ &amp;D) \dfrac{1}{2} \log 2 \end{align}$
98641
How do I calculate $\lim_{x\to 1}\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}$.  Please help me.  Thanks!
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?  $\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} &lt; \epsilon$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} &gt; \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \sqrt{n} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow n &gt; \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
62609
I cant use lhopital's rule here is my attempt:  Limit of $(\frac{n^{2}-2n+1}{n^{2}-4n+2})^{n}$ =$e^{\lim_{n\to \infty} n\times \ln(\frac{1 -2-n +\frac{1}{n^{2}}}{1 -\frac{4}{n} +\frac{2}{n^{2}}})}$ I got stuck here
51360
If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$  Find $\lim_{n \to \infty} nt_n$  First attempt: $t_n$ is positive(grouping two terms and performing subtraction we will get it) so is $nt_n$. Now can we prove it is monotonically decreasing? If so then $\lim_{n \to \infty} nt_n=\lim_{n \to \infty}[(\frac{1}{2+1/n}-\frac{1}{2+2/n})+(\frac{1}{2+3/n}-\frac{1}{2+4/n})+\cdots +(\frac{1}{4-1/n}-\frac{1}{4})]$ and each of these terms will go to zero so is the limit.  Second attempt: I was trying to use Riemann summation $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}= \frac{1}{2n+1}+\frac{1}{2n+2}+\frac{1}{2n+3}+\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}+\frac{1}{4n}-2[\frac{1}{2n+2}+\frac{1}{2n+4}+\cdots \frac{1}{4n}]\Rightarrow \lim \frac 1n [nt_n]=\int_0^2\frac{dx}{2+x}-\int_0^1\frac{dx}{1+x}=\ln4-\ln 2-\ln 2=0$  So $\lim t_n=0$  So what will happen with $\lim nt_n$  Edit: As I got the answer is not $0$ because of the flaw. So can we have different approaches even with Riemann Sum to have the answer?
105571
I guess it has something to do with Riemann sums but this is new for me.  $\displaystyle\lim \limits_{n \to \infty}\sum \limits_{k=n}^{2n}\frac{k}{k^2+n^2}$  How do I start?
4837
Let $$a_n=\left(1-\dfrac{1}{\sqrt2}\right)\dots \left(1-\dfrac{1}{\sqrt{n+1}}\right),n\ge1$$ Then find $\lim_{n\to \infty} a_n$.   How can I proceed? I am stuck at the first step. Please help.