Evaluate: $$\lim_{n\to \infty}\sqrt{n}\int_{0}^1 \frac{1}{(1+x^2)^{n}}dx$$
13215
$$  \lim_{n\to\infty} n^{1/2} \int_{0}^{1} \frac{1}{(1+x^2)^n}\mathrm{d}x=0 $$  Is my answer correct? But I am not sure of method by which I have done.
2355
Calculate $$\lim_{n \to \infty} \frac{2^3-1}{2^3+1}\times \frac{3^3-1}{3^3+1}\times \cdots \times\frac{n^3-1}{n^3+1}$$ No idea how to even start.
54106
$\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}$  Here's what I did:  $\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}\\ =(\lim(\frac94)^n+\lim((1+\frac1n)^{n})^n)^{\frac1n}\\ =(\lim(\frac94)^n+\lim e^n)^{\frac1n}\\$  Any hints on how to continue?  PS: no logs/integration/derivation because we haven't covered it.
101257
Could you explain how to find this limit?  $\lim_{n \rightarrow \infty} \sum^n_{k=1} \frac{1}{k+n}$
29097
Calculate below limit $$\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$$
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.
121419
How to prove the following: $$ \lim_{n \to \infty} \int_0^1 \int_0^1 \cdots \int_0^1 \frac{x_1^2+x_2^2+ \cdots +x_n^2}{x_1+x_2+ \cdots +x_n} dx_1 dx_2 \cdots dx_n = \frac23 $$  I would really appreciate if you could help me!
70484
Compute   $\sum_{n=1}^\infty \frac{1}{(n+1)\sqrt[]{n}+n\sqrt[]{n+1}}$
48990
What is the result of  $\lim_{(x,y)\to (0,0)}\frac{x+y}{\sqrt{x^2+y^2}}$ . I tried to do couple of algebraic manipulations, but I didn't reach to any conclusion.  Thanks a lot.
120997
$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$  Can we write it as following  $E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$  Let's see what happens:-  $$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$ $$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$  In the same way for further terms, we will get $0$  Let's also confirm for general term  $$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$ $$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$  So the whole expression $E$ will be zero  But actual answer is $1$  Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$  $OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$  Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?  I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.  But I got the following counter thought:-  $\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$  As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|&lt;1$  $\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$  $\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$  Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity.   But surprisingly $\dfrac{1}{3}$ is the correct answer.  I am feeling very confused in these two things. Please help me.
117296
Find $$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\ldots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}\right)$$
43463
Evaluate $$\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$$
107611
Why is $\lim_{x\to\infty} \frac{1}{x}$ equal to $0$?
23497
How do I evaluate this?  $$\lim_{n\to \infty }\sum_{k=1}^{n}\frac{k}{k^2+n^2}$$  I got concerned for that, I've tried make it integral for Riemann but it still undone.
113177
assume  $$x_n=\frac{n+1}{2^{n+1}}\sum_{k=1}^n\frac{2^k}{k} ,n=1,2,.....$$  how compute $\lim_{n\to +\infty}x_n$?  Thanks for any hint
40686
Let $f$ be $f\in C([0,\infty ])$, such that $\lim_{x \to \infty} f(x) = L $.  Calculate  $$ \int _{0}^\infty \frac{f(x)-f(2x)}{x}dx $$  Help?
99383
Find the limit,  $$L=\lim_{n\to \infty}\int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx$$  My try:  $$ \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}xdx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}(1-x)dx&lt; \int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx&lt; \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}(1-x)dx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}xdx$$  Now taking the limit I get that,  $$\frac{1}{4}&lt;L&lt;\frac{3}{4}$$  But, how can I get the exact answer!!  This is  from the American Mathematical Monthly.