I have this function which I need to calculate the limit of $\log(x)^{\log(x)}$ as $x$ approaches 1. While my mind tells me that with logic, $0+^0+$ equals to number to the power of 0, therefore = 1, but I don't know how to prove it.
120469
How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$ ? i know that its "1", but why? How can i calculate this?  Thank you very very much =)
91429
I have a midterm coming up and on the past exams the hard question(s) usually involve some form of $\lim_{x\to0^{+}} x \ln x$. However, we're not allowed to use l'Hopital's rule, on this year's exam anyways.  So how can I evaluate said limit without l'Hopital's rule? I got somewhere with another approach, don't know if it's useful:   $\lim_{x\to0^{+}} x \ln x = \lim_{x\to0^{+}} x^2 \ln (x^2) = L$ $= (\lim_{x\to0^{+}} 2x)(\lim_{x\to0^{+}} x \ln x)$ $= 0 * L$   Then I just need to prove that L is finite/exists (which means it must be 0)
19015
$$\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)$$  Now, $\log(n!) = \Theta (n\log(n))$ so I think we could write,   $$\lim_{n\to\infty}\frac{1}{2n}\left(\log\left(2n!\right) - \log\left(n!^2\right)\right) = \frac{1}{2n}\left(\log\left(2n!\right) - 2\log\left(n!\right)\right) $$  $$\lim_{n\to\infty}\frac{1}{2n}\left(2n\log\left(2n\right) - 2n\log\left(n\right)\right) $$  $$\log\left(2n\right) - \log\left(n\right) = \log(2)$$  Is this legitimate? I feel this might be wrong since, the $\Theta$ notation conceals a constant factor.
86379
I have to solve: $\lim\limits_{x\to \infty} \frac{\ln(x+1)}{\ln(x)}$. Can you give me any hints to go? Thanks a lot!
45090
I need to find the following:  $$\lim_{x\to\infty} f(x)= \lim_{x \to \infty}\left (\frac{x}{x+1} \right )^x$$  I know that this limit = $\frac{1}{e}$ from plugging it into a calculator, but I have to prove it without using the fact that:  $$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}\left (\frac{x}{x+k} \right )^x=\frac{1}{e^k}$$  I started by exponentiating:  $$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}e^{\ln \left (\frac{x}{x+1} \right )^x}$$  and from here I've dropped the exponent $x$ in front of the $\ln$, and from here I'm getting stuck. Should I separate the $\ln$ like this?  $$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}e^{x(\ln(x)-\ln(x+1))}$$  This doesn't seem to be leading me down the right path, but I'm not sure how else to do it. Is there a way to apply L'Hopital? If so, how?
107208
$\log x =Cx^4$ has only one root. Find C.  I don't know how to solve this problem. Do you take derivative on both sides?  I am thinking C equals 0. Am I correct on that?
14087
Find $\lim_\limits{x\to 1^-}{\ln x\cdot \ln(1-x)}$. I can't even start because I don't really know what $x\to 1^-$ means. If you know what it means it would really help me. I would as well if you helped me with finding the limit, hinted or thorough.
71361
$\lim\limits_{(x,y) \to (0,0)}\ {xy\log(x^2+y^2)}$  I tried this:  $$xy\log(x^2+y^2)=xy\log(x^2)+xy\log(1+y^2/x^2)$$  $|xy\log(x^2)|=2|x||y||\log|x||$ (because I use $(x,y)\neq(0,0)$) $\le 2|x||y|||x|-1|$ and taking the limit, $\lim_{(x,y) \to (0,0)}\ {xy\log(x^2)}=0$.  Since the other member has two variables in the argument of $\log$, I don't know how to continue. I know the limit of this function is zero because proceeding with other methods I've seen the limit is zero.
77476
I have to prove that      $$\Big[\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)\Big] \to \frac 12$$   I know that $$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)= \\ =\log\Big(\frac {n^2+1}{n^2}\Big)+ \log\Big(\frac {n^2+2}{n^2}\Big)+...+ \log\Big(\frac {n^2+n}{n^2}\Big)=\\ =\log\Big(n^2+1\Big)+ \log\Big(n^2+2\Big)+...+ \log\Big(n^2+n\Big)-2n\log n$$  Doing some manipulation we get:           $$\lim_{n \to +\infty}\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)=\frac 12$$      How can I prove this limit?
12062
I have no idea where to even start. WolframAlpha cant compute it either.  $$\int^1_0 \log^2(1-x) \log^2(x) \, dx$$  I think it can be done with series, but I am not sure, can someone help a little so I can get a start??  Thanks!
109013
How would I work out a limit of the form:  $$\lim_{x\to 0}\;(1+x)^{1/x}$$  I know these types of limits have a solution based on $e$ but how do I find this solution?
85412
How find the asymptotic behavior for $x(n)$  if $x^{x^x} = n$?  I supposed that $x = O(\log\log{n})$ and took logarithm two times.  So I get $x = O(\frac{\log\log{n}}{\log\log\log{n}})$  Is it right? How to improve this estimate?
50045
Use the defintion of a logarithmic function to prove $$ \frac{1}{x+1} \le \ln(1+\frac{1}{x}) \le \frac{1}{x}$$ for any positive integer x.   ( Using defintion : $e^y=\ln^{-1}y$ )     Attempt:    Using the definition,  $e^y=\ln^{-1}y$ ,    I tried to prove the right hand side of the equality first by $ 1 +\frac{1}{x} \gt \frac{1}{x}$,     $\ln(1 +\frac{1}{x})\lt \frac{1}{x} \Rightarrow   e^{\ln(\frac{x+1}{x})} \le e^{\frac{1}{x}}$ but I am really stuck and lost on how I should do it so that one of the sides there is ln and the other side is the value and also for the equality to be $\le$ as $ 1 +\frac{1}{x} \gt \frac{1}{x}$  is strictly less than and is not equal.
71433
How do you prove $\lim_{n\rightarrow\infty} n(2^{1/n} - 1) = \log 2$ ?  Background: in computer science, if you allocate CPU time to $n$ processes by , all the processes get sufficient amount of time when the quantity $U$ called CPU utilization is less than or equal to $n(2^{1/n} - 1)$.  It is monotonously decreasing and tends to $\log 2$ when $n\rightarrow\infty$, so if $U \le \log 2$, you can be sure all the processes will be given sufficient CPU time.
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$$ \int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2) $$  Anyone an idea on how to prove this?
23626
Evaluate $\displaystyle \lim_{x \to 0} \left(\frac{1}{\ln(1+x)} +  \frac{1}{\ln(1-x)}\right)$   I am having trouble starting this one. I couldn't see any log laws that I'm familiar with to rearrange the formula. I also tried combining the fractions and using L'Hospital's, but it only seemed to make things worse.  What direction should I take with this?
117031
$$\lim\limits_{x\to\infty} \ \frac{\ln \ x}{x}  = 0$$  How can I know this without looking at the graph? What's the easiest way to equate that expression to 0?  I understand that an integer over x approaches 0 as x approaches 0, but since ln x is an increasing function as x approaches infinity, how do we know which effect outweighs the other?