I have tried simplifying in every way I can think of, but I just can't seem to figure this out. I am asked to find the following limit: $$\lim_{n\to \infty} \sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}$$
122311
$$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$$  I tried to say we can erase the $1$ from the equation, as it's a constant. But I don't know how to do the rest without running into this mistake: $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-n=\frac{\sqrt[3]{\frac{n^3}{n^3}+\frac{n^2}{n^3}}-\frac{n}{n}}{\frac{1}{n}}=\frac{1-1}{0}$$
91998
I have seen this question on the internet and was interested to know the answer.  Here it is :  Calculate $\lim\limits_{n\to\infty}(1+\sqrt[2]{2+\sqrt[3]{3+\dotsb+\sqrt[n]n}})$?  Edit : I really tried doing it but wasn't able to get somewhere.  I know how to do questions like  $ y =  (1+\sqrt{1+\sqrt{1+\dotsb+\sqrt 1}}) $ and then we write $ (y-1)^2 = y $ and solve.   But for this I have no method. So I would like even a sort of a hint to try to get me started, no need for answer.
53852
Hi I have a question regarding finding the values of limit for the following question.  Let $a, b \in \mathbb R$. Find the limit $$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$
32524
How can I find the $\lim  (1+(a^{1/n}-1) / b)^n$ when $n \to \infty$?    I would greatly appreciate it if you kindly give me some hints.
115805
$\lim_{n\to \infty}\sqrt[n]{3^n+5^n}$  I'm stuck. Tried to apply the squeeze theorem, but found only the $\le$ side, which is $\sqrt[n]{5^n}$, approaching 5.  How do I proceed from there?
42353
I’m having a problem approaching the question. The fact that it’s a limit of a sum is confusing to me.   $$\lim _{n\to \infty }\left(\frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}\right)$$
40900
I got a limit:  $$\lim_{x\to\infty}\frac {(2x+3)^3(3x-2)^2} {(x^5 + 5)}$$  As far as $x$ approaches infinity, can I just forget about 'small' numbers (like $3$, $-2$ and $5$ in this example)? I mean is it legal to make a transition to:  $$\lim_{x\to\infty}\frac {(2x)^3(3x)^2} {x^5}$$  Or if it is not always okay — in what cases such transitions are okay?
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
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!
21647
I need some suggestions, hints for the limit when $a \to 1^{-}$ of $$\sqrt{\,1 - a\,}\,\sum_{n = 0}^{\infty}a^{n^{2}}.$$
60294
How to solve the following limit question? $$\lim_{n\to\infty}\frac{1}{e^n}\Bigl(1+\frac1n\Bigr)^{n^2}$$  Thanks a lot.
112932
Any hints on how to find the following limit ?.  I haven't been able to figure it out still. $$ \lim_{n \to \infty}\left(\, 1 - \frac{1}{3}\, \right)^{2} \left(\, 1 - \frac{1}{6}\, \right)^{2}\left(\, 1 - \frac{1}{10}\, \right)^{2}\ldots\left[\, 1 - \frac{1}{n\left(\, n + 1\, \right)/2}\, \right]^{2} $$  Pardon me if it's a silly question.  Edit: Solved upto this much   $$ \prod_{n = 2}^{\infty}\left[\, 1-\frac{2}{n\left(\, n + 1\, \right)}\, \right] = \prod_{n = 2}^{\infty}\left(\, \frac{n + 2}{n + 1}\,\frac{n - 1}{n}\, \right) $$  I did'nt notice most of the terms cancel out.
59390
Let $x&gt;0$, find the limit $$\lim_\limits{n\to{\infty}}{n^2\left(\sqrt[n]{x}-\sqrt[{n + 1}]{x}\right)}$$  I use Maclaurin series and find out that the limit is $\ln x$. And this is the answer I get from a math forum:  "Let $x_n=\sqrt[n]{x}-1$, $y_n=\dfrac{1}{n}$, then $(x_n)\to0,(y_n)\to0,(y_n)\downarrow$ when $n\to\infty$.  Use Stolz-Cesaro theorem, we have:  $$\mathop {\lim }\limits_{n \to \infty } \dfrac{{\sqrt[n]{x} - \sqrt[{n + 1}]{x}}}{{\frac{1}{n} - \frac{1}{{n + 1}}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\sqrt[n]{x} - 1}}{{\frac{1}{n}}} = \ln x$$ Then $\mathop {\lim }\limits_{n \to \infty } {n^2}\left( {\sqrt[n]{x} - \sqrt[{n + 1}]{x}} \right) = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{n^2}}}{{n\left( {n + 1} \right)}}\cdot\dfrac{{\sqrt[n]{x} - \sqrt[{n + 1}]{x}}}{{\frac{1}{n} - \frac{1}{{n + 1}}}} = \ln x$.  /END"  I just wonder what form Stolz-Cesaro theorem that he used, because I just learn Stolz-Cesaro theorem when $(y_n)\to\infty$.  Could anyone help me to get this? Thank you in advance.     This is "another form of Stolz theorem" in the forum that I mentioned: Let $(x_n),(y_n)$     be two sequences of real numbers. Assume $\mathop {\lim }\limits_{n \to \infty } {x_n} = \mathop {\lim }\limits_{n \to \infty } {y_n} = 0$; $y_n&gt;0$, $(y_n)$ is strictly decreasing  and the following limit exist $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{x_{n + 1}} - {x_n}}}{{{y_{n + 1}} - {y_n}}} = L$. Then $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{x_n}}}{{{y_n}}} = L$. Is this correct?
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?   $$\lim_{n\to\infty} \left[ n-{n\over e}\left(1+{1\over n}\right)^n \right].$$
73834
Find the limit of: $$\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n$$   I'm pretty sure it goes to zero since $(n+1)^n &gt; n^n$ but when I input large numbers it goes to $0.36$.  Also, when factoring: $$n^{1/n}\left(\frac{1}{1+\frac1n}\right)^n$$ it looks like it goes to $1$.   So how can I find this limit?
59758
Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$  I did the following:  \begin{align} (\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})} \end{align}  I know the final answer is $\frac{9}{2}$. After multiplying by the conjugate, I see where the $9$ in the numerator comes from. I just can't remember how I solved the rest of the problem.
74371
I want to show that $$\lim_{n \to \infty} \sqrt{n}(\sqrt[n]{n}-1) = 0$$ and my assistant teacher gave me the hint to find a proper estimate for $\sqrt[n]{n}-1$ in order to do this. I know how one shows that $\lim_{n \to \infty} \sqrt[n]{n} = 1$, to do this we can write $\sqrt[n]{n} = 1+x_n$, raise both sides to the n-th power and then use the binomial theorem (or to be more specific: the term to the second power). However, I don't see how this or any other trivial term (i.e. the first or the n-th) could be used here.   What estimate am I supposed to find or is there even a simpler way to show this limit?  Thanks for any answers in advance.