I'm trying to solve this limit, for which I already know the solution thanks to Wolfram|Alpha to be $\sqrt[3]{abc}$:  $$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n:\forall a,b,c\in\mathbb{R}^+$$  As this limit is an indeterminate form of the type $1^\infty$, I've been trying to approach it by doing:  $$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}\cdot\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}=e^{\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}$$  But now when I approach that top limit this is what I get:  $$\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n=\lim_{n\rightarrow\infty}\frac{n\cdot a^{\frac{1}{n}}}{3}+\frac{n\cdot b^{\frac{1}{n}}}{3}+\frac{n\cdot c^{\frac{1}{n}}}{3}-n=\lim_{n\rightarrow\infty}\frac{n\cdot a^0}{3}+\frac{n\cdot b^0}{3}+\frac{n\cdot c^0}{3}-n=\lim_{n\rightarrow\infty}\frac{n}{3}+\frac{n}{3}+\frac{n}{3}-n=0$$  And hence the final limit should be $e^0=1$ which is clearly wrong but I honestly don't know what I did wrong, so what do you suggest me to solve this limit?
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 ^^
87800
I know what solution is $e^a$ but I don't know how to calculate this limit:  $$\lim \limits_{n \to \infty}\left(1 + \frac{a}{n}\right)^n=e^a$$  Can someone explain the steps to me?
67590
I need to find:      $$ \lim_{n \to \infty } (\sqrt{2}-\sqrt[3]{2})(\sqrt{2}-\sqrt[4]{2})(\sqrt{2}-\sqrt[5]{2})\cdot \cdot \cdot (\sqrt{2}-\sqrt[n]{2})  $$   So far, I think that $0&lt;\sqrt{2}-\sqrt[n]{2}&lt;1$, and it seems to me that the limit will approach zero but I can't figure how to show it mathematically.
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?
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.
3821
Can you help me with this limit? What do I have to do? I'm lost.  $$\lim_{n\to\infty}n\left(\sum_{i=1}^{n}\dfrac{1}{(n+i)^2}\right)$$  The solution given is $\dfrac{1}{2}$.
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!
94944
Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\frac{n!\cdot e^n}{\sqrt{n}\cdot n^n}$   $\bf{My\; Try::}$we can write it as $$l=\lim_{n\rightarrow \infty}\frac{e^n}{\sqrt{n}}\cdot \left(\frac{1}{n}\cdot \frac{2}{n}\cdot \frac{3}{n}\cdots \cdots \frac{n}{n}\right)$$  $$\ln (l) = \lim_{n\rightarrow \infty}\bigg[n-\frac{1}{2}n+\sum^{n}_{r=1}\ln\left(\frac{r}{n}\right)\bigg]$$  Now how can i solve it, Help required, Thanks
50671
I was trying to solve a problem and got stuck at the following step:  Suppose ${n \to \infty}$ .  $$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = 1$$  Let us rewrite $n^3=n \cdot n^2$ as $n^2 + n^2 + n^2 + n^2 \dots +n^2$,$\space$ n times.  Now we have  $$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = \frac {n^2 + n^2 + n^2 + n^2 + n^2 \dots +n^2}{n^3} $$  As far as I understand, we can always rewrite the limit of a sum as the sum of limits ...  $$\dots = \lim \limits_{n \to \infty} \left(\frac{n^2}{n^3} + \frac{n^2}{n^3} + \dots + \frac{n^2}{n^3}\right)$$  ...but we can only let ${n \to \infty}$ and calculate the limit if all of the individual limits are of defined form (is this correct?). That would be the case here, so we have:  $= \dots \lim \limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n}\right) =$[ letting ${n \to \infty}]$ $= 0 + 0 + \dots + 0 = 0$  and the results we get are not the same.  Where did I go wrong?
2244
I'm trying to find  $$\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} .$$   I tried couple of methods: Stolz, Squeeze, D'Alambert  Thanks!  Edit: I can't use Stirling.
90576
$\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$  is equal to :  $\frac{9}{e^2}$ $3 \log3−2$ $\frac{18}{e^4}$ $\frac{27}{e^2}$    My attempt : $\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$ $=\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(n+2n)}{n^{2n}})^{\frac{1}{n}}$ $=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(1+2n/n)\}}{n^{2n}})^{\frac{1}{n}}$ $=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(1+2)\}}{n^{2n}})^{\frac{1}{n}}$ $=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(3)\}}{n^{2n}})^{\frac{1}{n}}$ $=\lim_{n \to \infty} (\{(1+1/n)(1+2/n)\dots(3)\})^{\frac{1}{n}}$ I'm stuck here.  Can you please explain?
84645
I need to solve   $$\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{x-1} $$  But without using L’Hopital’s Rule, could someone help me?
70716
Consider this limit:  $$ \lim_{x \to 1} \frac{\sqrt x -  1}{ \sqrt[3] x -  1}  $$  The answer is given to be 2 in the textbook.  Our math professor skipped this question telling us it is not in our syllabus, but how can it be solved?
91070
I'm struggling to find the limit at infinity of : $$\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}), n\in\Bbb N$$ I know it is $1$ but I don't understand why this is wrong :  $\sin^2(\pi\sqrt{n^2 + n}) = \sin^2(\pi*n\sqrt{1 + \frac{1}{n}})$.  So $\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}) = \lim \limits_{n\rightarrow\infty}\sin^2(\pi*n\sqrt{1 + \frac{1}{n}}) =  \lim \limits_{n\rightarrow\infty} \sin^2(\pi*n) = 0$.  Thanks.
100248
how I can continue this limit resolution?  The limit is:   $$ \lim_{n \to \infty}{\sqrt[n]{\frac{n!}{n^n}}} $$  This is that I have done:  I apply this test: $ \lim_{n \to \infty}{\sqrt[n]{a_n}} = \frac{a_{n+1}}{a_n} $  Operating and simplifying I arrive to this point: $$ \lim_{n \to \infty}{\frac{n^n}{(n+1)^n}} $$  I've done something wrong? Thanks!
93101
Find the limit of the sequence {$a_{n}$}, given by$$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{n+1}=\dfrac {1}{3}(1+a_{n}+a^{3}_{n-1}), \ for \ n \ &gt; \ 1$$  My try:  $ a_{1}=0,a_{2}=\dfrac {1}{2},a_{3}=\dfrac {1}{2},a_{4}=0.54$ that is the sequence is incresing and each term is positive. Let the limit of the sequence be $x$. Then $ \lim _{n\rightarrow \infty }a_{n+1}=\lim _{n\rightarrow \infty }a_{n}=x$ $$ \lim _{n\rightarrow \infty }a_{n+1}= \lim _{n\rightarrow \infty }1+a_{n}+a^{3}_{n-1}$$  $\Rightarrow x=\dfrac {1}{3}( 1+x+x^3)$  $\Rightarrow x^3-2x+1=0$  and this equation has three roots $x=\dfrac {-1\pm \sqrt {5}}{2},1$   So the limit of the sequence is $\dfrac {-1 + \sqrt {5}}{2}$.  how can i say that the limit is $\dfrac {-1 + \sqrt {5}}{2}$?
91502
Find the limit $$\lim_{x \to \infty}\left((x+3)^{1+1/x}-x^{1+1/(x+3)}\right).$$   I did nothing by now.