1
$\begingroup$

I have tried to evaluate this limit:

$$\lim_{x\to \infty } \left(x(x+1) \log \left(\dfrac{x+1}{x} \right)-x\right)=\frac12$$

using $\lim_ {x\to \infty }\left(1+\dfrac{1}{x}\right)^{x}=e$, and using the variable change $z=\dfrac{1}{x}$ to get some known and standrad limit related to $\log$ natural logarithm properties function but I didn't succeed? Then any way and it's good if there is a suitable way for high school level.

$\endgroup$
  • $\begingroup$ $$\log (\frac{x+1}{x} )=\log (1+\frac{1}{x} )= \frac{1}{x} -\frac{1}{2x^2} +\frac{1}{3x^3} - \ldots$$ when $x>1$ $\endgroup$ – Yuriy S Aug 17 at 19:09
  • $\begingroup$ Thanks , but is there anyway available for high school student $\endgroup$ – zeraoulia rafik Aug 17 at 19:11
  • $\begingroup$ But if you are not allowed to use Taylor series (an unfortunate, but real occurence) then I can try to work it out in another way $\endgroup$ – Yuriy S Aug 17 at 19:15
  • $\begingroup$ Thanks youris , I want a sutiable way for high school level if it is possible , i have tried manytimes but no result $\endgroup$ – zeraoulia rafik Aug 17 at 19:16
  • 1
    $\begingroup$ Can you use L'Hospital's Rule? $\endgroup$ – DonAntonio Aug 17 at 19:21
3
$\begingroup$

Further to my comment, L'Hôpital's rule applied twice $$\lim\limits_{x\to\infty}\frac{(x+1)\log\left(\frac{x+1}{x}\right)-1}{\frac{1}{x}}= \lim\limits_{x\to\infty}\frac{\log\left(1 + \frac{1}{x}\right)-\frac{1}{x}}{-\frac{1}{x^2}}=\\ \lim\limits_{x\to\infty}\frac{\frac{1}{x^2 + x^3}}{\frac{2}{x^3}}= \lim\limits_{x\to\infty}\frac{1}{2}\cdot\frac{x^3}{x^2+x^3}=\frac{1}{2}$$ It is worth mentioning that both times we are dealing with $\frac{0}{0}$, so L'Hôpital's rule can be applied. L'Hôpital's rule used to be part of the high school program, I hope it still is.

$\endgroup$
1
$\begingroup$

Set $1/x=h\implies h\to0$

$$\lim_h\dfrac{(1+h)\ln(1+h)-h}{h^2}=\lim_h\dfrac{\ln(1+h)-h}h+\lim_h\dfrac{\ln(1+h)}h$$

For the first limit, use Are all limits solvable without L'Hôpital Rule or Series Expansion

What about the second one?

$\endgroup$

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.