1
$\begingroup$

How to compute $$\lim_{x\to +\infty}\frac{(x+1)^x}{x^{x+1}}$$ I'm interested in more ways of computing limit for this expression

My Thoughts $$\lim_{x\to +\infty}\frac{(x+1)^x}{x^{x+1}}=0$$ indeed, \begin{align} \lim_{x\to +\infty}\frac{(x+1)^x}{x^{x+1}}&=\lim_{x\to +\infty}\exp\left( \ln\left( \dfrac{(x+1)^x}{x^{x+1}} \right)\right)\\ &=\lim_{x\to +\infty}\exp\left( \ln\left( (x+1)^x.x^{-(1+x)} \right)\right)\\ &=\lim_{x\to +\infty}\exp\left( \ln\left( (x+1)^x \right)+\ln\left(x^{-(1+x)}\right) \right)\\ &=\lim_{x\to +\infty}\exp\biggl(x.\ln\left( x+1 \right)+\left(-1-x\right).\ln(x)\biggr)\\ &=\lim_{x\to +\infty}\exp\biggl(x.\ln\left(x(1+\dfrac{1}{x}) \right)+\left(-1-x\right).\ln(x)\biggr)\\ &=\lim_{x\to +\infty}\exp\biggl(x.\ln(x) +x\ln\left(1+\dfrac{1}{x} \right)+\left(-1-x\right).\ln(x)\biggr)\\ &=\lim_{x\to +\infty}\exp\biggl(x\ln\left(1+\dfrac{1}{x} \right)-\ln(x)\biggr)\\ &=\lim_{x\to +\infty}\exp\biggl(\dfrac{\ln\left(1+\dfrac{1}{x} \right)}{\dfrac{1}{x}}-\ln(x)\biggr)=0\\ \end{align} since $\lim_{x\to +\infty}\dfrac{\ln\left(1+\dfrac{1}{x} \right)}{\dfrac{1}{x}}=1$

  • Is my proof correct
$\endgroup$
  • $\begingroup$ what is the $\lim_{x \to \infty}\frac{x^x}{x^{x+1}}?$ $\endgroup$ – abel Jan 18 '16 at 17:36
  • $\begingroup$ @abel $\lim_{x \to \infty}\frac{x^x}{x^{x+1}}=\lim_{x \to \infty}\frac{1}{x}=0$ $\endgroup$ – Educ Jan 18 '16 at 17:49
9
$\begingroup$

Write $$ \frac{(x+1)^x}{x^{x+1}}= \frac{1}{x}\left(\frac{x+1}{x}\right)^{\!x}= \frac{1}{x}\left(1+\frac{1}{x}\right)^{\!x} $$ Then…

Alternative way: let's compute the limit of the logarithm: \begin{align} \lim_{x\to\infty}(x\log(x+1)-(x+1)\log x) &= \lim_{t\to0^+}\frac{1}{t}\log\left(\frac{1}{t}+1\right)-\left(\frac{1}{t}+1\right)\log(1/t)\\ &=\lim_{t\to0^+}\frac{\log(1+t)+t\log t}{t}\\ &=\lim_{t\to0^+}\left(\frac{\log(1+t)}{t}+\log t\right)\\ &=-\infty \end{align} Note that this is basically your own solution. But the substitution $x=1/t$ makes it less heavy, in my opinion.

$\endgroup$
1
$\begingroup$

I would say everything looks right. This is a lengthy way to solve the problem, but since you simply ask if the proof is correct, the answer is yes. Now, your last line (the one equating to zero) involves breaking up the limit into two parts and then taking the limit of $\frac 1x$ as $x \to \infty$. While this is true, you might want to split up the multiple steps by adding a line or two there. There are special conditions that allow you to bring the limit inside of the exponential and logarithms, but both satisfy these criteria so you are fine... Just be aware this does not hold in general with all functions and their inverse.

$\endgroup$
  • $\begingroup$ so my functions are continuous that's why i have right to bring limit nside of the exponential and logarithms? am i wrong $\endgroup$ – Educ Jan 18 '16 at 17:46
  • $\begingroup$ @Educ no, that's pretty much right :) you can go a little deeper than that, but you're right... That condition should suffice here! $\endgroup$ – Brevan Ellefsen Jan 18 '16 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.