$$\lim _{ x\rightarrow { 1 }^{ - } }{ \prod _{ n=0 }^{ \infty  }{ { \left( \frac { 1+{ x }^{ n+1 } }{ 1+{ x }^{ n } }  \right)  }^{ { x }^{ n } } }  } $$ I tried it as follow but not able to proceed further $$L=\lim _{ x\rightarrow { 1 }^{ - } }{ \prod _{ n=0 }^{ \infty  }{ { \left( \frac { 1+{ x }^{ n+1 } }{ 1+{ x }^{ n } }  \right)  }^{ { x }^{ n } } }  } \\ \ln { L } =\lim _{ x\rightarrow { 1 }^{ - } }{ \sum _{ n=0 }^{ \infty  }{ \ln { { \left( \frac { 1+{ x }^{ n+1 } }{ 1+{ x }^{ n } }  \right)  }^{ { x }^{ n } } }  }  } \\ \ln { L } =\lim _{ x\rightarrow { 1 }^{ - } }{ { x }^{ n } } \sum _{ n=0 }^{ \infty  }{ \ln { { \left( 1+\frac { { x }^{ n }(x-1) }{ 1+{ x }^{ n } }  \right)  } }  } \\ \ln { L } \le \lim _{ x\rightarrow { 1 }^{ - } }{ (x-1)\left( \frac { { x }^{ 2 } }{ 1+x } +\frac { { x }^{ 4 } }{ { 1+x }^{ 2 } } +\frac { { x }^{ 6 } }{ { 1+x }^{ 3 } } +..... \right)  } $$
5291
Compute : $$L=\lim_{x\to 1^-} \left(\prod_{n=0}^{\infty} \left(\frac {1+x^{n+1}}{1+x^n}\right)^{x^n}\right)$$   My try:   Writing out the first few terms I noticed that the limit can be expressed as $$L=\lim_{x\to 1^-} \frac 12\left(\prod_{n=1}^{\infty} (1+x^n)^{x^{n-1} -x^n}\right)$$ $$L=\frac 12\lim_{x\to 1^-} \left(\prod_{n=1}^{\infty} (1+x^n)^{x^{n-1}}\right)^{1-x}$$  Converting to exponential form I get $$L=\frac 12 \exp {\left(\lim_{x\to 1^-} (1-x)\sum_{n=1}^{\infty} \left(x^{n-1}\ln(1+x^n)\right)\right)}$$  And got stuck here. I did try to use approximation $\ln x\sim x$ hereon to get final answer as $\frac {\sqrt e}{2}$ but I think it was improper to use approximation and hence I believe that the answer I got is flawed. I also noticed that the form I obtained is quite similar to be used for a Riemann Sum but I didn't get a way to tackle the problem using that way.  Can someone please help me with this problem....
108876
Let a sequence be $a_1=1;a_{n+1}=(n+1)(1+a_n)$ If $P_n=\prod_1^n(1+a_i^{-1})$ then $$\lim_{n\to\infty}P_n $$is ?    I did: $$P_n=\prod_1^n\frac{(1+a_i)}{a_i} =\frac{a_{n+1}}{a_1}\prod1^{n-1}\frac{(1+a_i)}{a_{i+1}} =\frac{a_{n+1}}{a_1}\prod_1^{n-1}\frac1{i+1}=\frac{a_{n+1}}{n!}$$ Now how do I find the limit?
110106
x is a real number between 0 and 1  $ \lim_{n\rightarrow\infty} x ^ {\left(\frac 1 n\right)} = 1 $  SO far i know you need to do this to prove it:  let b>1, prove that $\root n \of b \rightarrow 1$ as $n \rightarrow\infty$  I was given this hint: let $a_n = \root n \of b - 1$ , prove that  $a_n \rightarrow 0$  Thanks!
95156
Determine the following limit $$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{(n+1)(n+2)\ldots (n+n)} $$ Idea #1 the nth root could be rewritten as $$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{\frac{(n+n)!}{n!}} $$ Noticing that $n^{-1}$ is an infinitesimal, if the n-th root were bounded, the limit would be $0$.  But hoping that $\sqrt[n]{\frac{(n+n)!}{n!}}$ is bounded is a pipe dream, this clearly tends to $\infty$.  Idea #2 $$\lim_{n\to\infty}\ln\left (\frac{1}{n}[(n+1)\ldots (n+n)]^\frac{1}{n}\right ) = \lim_{n\to\infty}\left [-\ln n+\frac{1}{n}\left (\ln (n+1)+\ln (n+2)+\ldots +\ln (n+n) \right )\right ]$$ but trying to bring it to a common denominator doesn't help either, what I would have is $$\lim_{n\to\infty}\frac{1}{n} \left [-n\ln n+\ln (n+1)+\ldots +\ln 2+\ln n\right ]$$ Which doesn't bring me any closer to simplification. The solution of this must something entirely different. I would like a hint/point me in the direction I should go.
102634
Evaluate:  $$ L=\lim_{n \to \infty} \left(\frac 1{n^2+1}+\frac 2{n^2+2}+\frac 3{n^2+3}+\cdots +\frac n{n^2+n}\right)$$  My approach:  Each term can be written as  $$ \frac k{n^2+k}=\frac {n^2+k-n^2}{n^2+k}=1-\frac {n^2}{n^2+k}$$  $$ \therefore \lim_{n \to \infty}\frac k{n^2+k}=\lim_{n \to \infty}\left(1-\frac {n^2}{n^2+k}\right)=0$$   hence,  $$ L=0$$  Problem:  The correct answer is 1/2, please indicate the flaw in my approach or post a new solution.  Thank You
44975
How may one calculate  $$\lim_{n\to\infty} \ \left(\left(\sum_{k=1}^{n} \frac{1}{3k-1}\right) - \frac{\ln n}{3}\right) \ ?$$
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.
42353
I’m having a problem approaching the question. The fact that it’s a limit of a sum is confusing to me.   $$\lim _{n\to \infty }\left(\frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}\right)$$
9283
How can I compute the limit $\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n^2}$? Of course $\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n} = e$, and then $\left(1+\frac{1}{n}\right)^{n^2} = \left(\left(1+\frac{1}{n}\right)^n\right)^n$. Since the term inside converges to $e$, the whole thing is $e^n$, and can I conclude that the limit is infinity as $n\rightarrow\infty$?
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: $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-n=\frac{\sqrt[3]{\frac{n^3}{n^3}+\frac{n^2}{n^3}}-\frac{n}{n}}{\frac{1}{n}}=\frac{1-1}{0}$$
3706
Evaluate the infinite product $$\lim_{ n\rightarrow\infty }\prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right ).$$  I can't see anything in this limit , so help me please.
91185
$$ \lim_{t\to 1^-} (1-t) \sum_{n=0}^\infty \frac{t^n}{1+t^n} $$  I try to use L'hopital's rule but not successfully. Also I know that exist formula Sonine but I can't understand how I can use it.
12735
$$\lim_{x\to 0} \frac{\ln(1+x)}x$$  The process I  want to take to solving this is by using the definition of the limit, but I am getting confused. ( without l'hopitals rule)   $$\lim_{h \to 0} \frac{f(x+h) - f(x)}h$$  $$\lim_{h \to 0} \frac{\frac{\ln (1+x+h)}{x+h} - \frac{\ln(1+x)}x}h$$  $$\lim_{h \to 0} \frac{x\ln(1+x+h) - (x+h)\ln (1+x)}{hx(x+h))}$$  At this point I get confused because I know the answer is $1$, but I am not getting this answer through simplification of my formula.
58710
What would be the nicest proof of the following theorem:  If $\lim\limits_{n \rightarrow \infty} a_{n} = \infty$, then $\lim\limits_{n \rightarrow \infty} \left(1 + \frac{1}{a_{n}} \right) ^ {a_{n} } = e$. If $\lim\limits_{n \rightarrow \infty}b_{n} = 0$, then $\lim\limits_{n \rightarrow \infty} \left(1 + b_{n} \right) ^ {\frac {1} {b_{n}} } = e$.  I somehow failed to find a proof here on the website and in the literature.
121778
$$\lim_{x\rightarrow 1} \frac{x^m-1}{x^n-1}$$   I found this in a book (recommended by the professor) without solutions or hints.  My problem is that I can't find definitions about this kind of problems and I have no idea how to start, any kind of help would be great. So far I've been dealing with "$\lim_{x\rightarrow ∞}$" only. Also without L'Hospital, because we haven't had this in lecture.
11141
Direct way leads to nowhere.  $$\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} } = \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{n \sqrt {1 +\frac{i}{n^2}} } = $$  $$ = \lim_{n\rightarrow \infty} \frac{1}{n} \cdot \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {1 +\frac{i}{n^2}} } = 0 \cdot \sum^{n}_{i=1} 1 = 0 \cdot n $$  So, I tried to apply squeeze theorem:  $$\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} } \leq \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{n}  = 1$$  Could you help me to figure out the lower bound?
124283
Show that $$\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{6n}\right)=\log 6$$ Here I need to use the definition of integral but I  faced problem in range . Please help.