I am interested in the limit     $$ \lim_{n \rightarrow \infty} \sqrt[n]{n^3}$$   Can we simply conclude that: $$ \lim_{n \rightarrow \infty} (\sqrt[n]{n})^3= 1^3=1.$$ I have proven that $\sqrt[n]{n}\rightarrow1$ earlier in this textbook. Also since the limit of a power is the power of the limit.
50511
Thomson et al. provide a proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$ in . It has to do with using an inequality that relies on the binomial theorem:  I have an alternative proof that I know (from elsewhere) as follows.  Proof. \begin{align} \lim_{n\rightarrow \infty} \frac{ \log n}{n} = 0 \end{align} Then using this, I can instead prove: \begin{align} \lim_{n\rightarrow \infty} \sqrt[n]{n} &amp;= \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \newline &amp; = \exp{0} \newline &amp; = 1 \end{align}  On the one hand, it seems like a valid proof to me. On the other hand, I know I should be careful with infinite sequences. The step I'm most unsure of is: \begin{align} \lim_{n\rightarrow \infty} \sqrt[n]{n} = \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \end{align} I know such an identity would hold for bounded $n$ but I'm not sure I can use this identity when $n\rightarrow \infty$. Question: If I am correct, then would there be any cases where I would be wrong? Specifically, given any sequence $x_n$, can I always assume: \begin{align} \lim_{n\rightarrow \infty} x_n = \lim_{n\rightarrow \infty} \exp(\log x_n) \end{align} Or are there sequences that invalidate that identity?  (Edited to expand the last question) given any sequence $x_n$, can I always assume: \begin{align} \lim_{n\rightarrow \infty} x_n &amp;=  \exp(\log \lim_{n\rightarrow \infty} x_n) \newline &amp;=  \exp(\lim_{n\rightarrow \infty} \log x_n) \newline &amp;=  \lim_{n\rightarrow \infty} \exp( \log x_n) \end{align} Or are there sequences that invalidate any of the above identities? (Edited to repurpose this question). Please also feel free to add different proofs of $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$.
90860
We know that $\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n=e$ and so I thought the approach to evaluating the limit in the question would be to just use this fact and substitute it into the numerator. This approach would tell us the above limit evaluates to $1$. However, that does not seem to be the correct limiting value. In fact, it evaluates to $\frac {1}{\sqrt{e}}$. Why is this so?
48433
Hi I have a question regarding of limits to infinity please help! Thank You!  The question states the user to find the following limit:  $  \lim_{n\to\infty} n^2 ({\sqrt[n]{x}-\sqrt[n+1]{x}}) $      Where $(x &gt; 0) $  Thank You!!!
39852
I'm having quite hard time proving the following statement: $$\lim_{n\to\infty}\sqrt[n]{2^n-n^2}=2$$  I though about using the squeeze theorem. So I started to look for upper and lower expressions and quickly found an upper limit to the expression,  $$\sqrt[n]{2^n-n^2}&lt;\sqrt[n]{2^n}=2$$  However, I found it very difficult to find the lower limit. Help would be appreciated!.
5287
Let $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}}$  a) Show that $x_{n} &lt; x_{n+1}$  b) Show that  $x_{n+1}^{2} \leq 1+ \sqrt{2} x_{n}$  Hint : Square $x_{n+1}$ and factor a 2 out of the square root  c) Hence Show that  $x_{n}$ is bounded above by 2. Deduce that $\lim\limits_{n\to \infty} x_{n}$ exists.  Any help? I don't know where to start.
38578
Prove that:  $$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{2^{n}}=0$$  I think I need to show $2^n \geq n^3$ $\forall n \geq10$. Is it true?  What about: $$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{n!}=0$$  What to do with $n!$?
49258
The limit is   $$\lim_{x\rightarrow0} \frac{x-\sin_n(x)}{x^3},$$  where $\sin_n(x)$ is the $\sin(x)$ function composed with itself $n$ times:  $$\sin_n(x) = \sin(\sin(\dots \sin(x)))$$  For $n=1$ the limit is $\frac{1}{6}$, $n=2$, the limit is $\frac{1}{3}$ and so on.  Can we define a recurrent relation upon that given hypothesis? Also, how do I involve $n$ into calculation, because the final limit will depend on it?  Any suggestions on how to tackle this?  Thank you!
91488
Let $a\geq 0$ and $ b\geq 0$. Prove that $\lim_{n \to \infty} \sqrt[n]{a^n+b^n}=\max \{a,b\}$. [Hint: Use the identity $(a^n -b^n)=(a-b)(\sum_{i=0}^{n-1}a^ib^{n-1-i})$]    I need some help! I cannot do it even with the hint... :(
109274
So recently a friend asked me to compute  limit:   $$\lim_{n\to \infty }\left(\sqrt[\leftroot{-2}\uproot{2}n+1]{(n+1)!}-\sqrt[\leftroot{-2}\uproot{2}n]{n!}\right)$$  Question : Does the limit exist? If yes is it finite and if yes what is its value?  How do we solve this?  Edit:  Note: I am only familiar with only basics of limit solving(upto  L'Hôpital's rule) and have reasons to believe that this limit can be solved using these methods. If you could keep your answer simple that should help.  Update  Here is where I have gotten so far   $$ \lim_{n\to \infty} (n+1)! ^{1\over n+1} - (n)! ^{1\over n}$$  Can be written as $$ \lim_{n\to \infty}[1*2*3*...(n+1)]^{1\over n+1}  - [1*2*3*...n] ^{1\over n}$$   $$\implies \lim_{n\to \infty} [(n+1)[{1 \over n+1}* {2 \over n+1}  * {3\over n+1}...* {n+1 \over n+1}]^{1 \over n+1} - (n)[{1 \over n}* {2 \over n}  * {3\over n}...* {n\over n}]^{1 \over n} ]$$  (Factoring n+1 out of first expression and n from second.)  $$\implies  \lim_{n\to \infty} [(n+1) e^{{1 \over n+1} (\sum_{r=1}^{n+1}ln({r\over n+1}))} - (n) e^{{1 \over n} (\sum_{r=1}^{n}ln({r\over n}))}  ]$$  From here I think second limit can be solved as a integral(limit of a sum) but I cannot solve first. How can I proceed further?  Thanks!
65279
How can I calculate this limit? $$\lim_{n\rightarrow\infty} \frac{7^{\sqrt{n+1}-\sqrt{n}}\cdot(\frac{n+1}{2})!\cdot(\frac{n+1}{2})!}{(n+1)\cdot(\frac{n}{2})!\cdot(\frac{n}{2})!}$$ I don't have idea and I will be happy for help.
48528
I want to prove the following limit:  $$\lim_{n \to \infty} \sqrt[n+1\;] {(n+1)!} - \sqrt[n] {(n)!} = \frac{1}{e}.$$  I searched the forum &amp; found the link here:  .  But still, there is no way out of the problem. So, please solve it.
19929
I have a problem with the calculation of the following limit. \begin{equation} \lim_{n\rightarrow \infty} \frac{1+\sqrt{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n} \end{equation} I do not know where to start! Thank you very much
20697
I am trying to find the limit of:  $$\lim_{n \to \infty} n(\sqrt[n]{2} - 1)$$  I know it should be very simple but I don't seem to get it. Thanks in advance for any help!
5154
What is the result of the next limit:   $$ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} $$  Why (theorem)?
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.
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 $\dfrac {-1 + \sqrt {5}}{2}$?
53287
I need to evaluate the limit $$\lim_{n\rightarrow \infty}(n-\sqrt{n+\alpha}\sqrt{n+\beta})$$  My work: $$(n-\sqrt{n+\alpha}\sqrt{n+\beta})=n-\sqrt{n^2+\alpha n+\beta n+\alpha\beta}$$ I understand that both $\alpha$ and $\beta$ are just constants. But, now how should I factor this? I understand the calculus behind limits, but the algebra troubles me.