Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to improve my counting limits. I've found some difficult examples:

  1. $\displaystyle\lim_{x \to +\infty}\left((x+1)^{1+\frac1x}-x^{1+\frac{1}{x+a}}\right)$

  2. $\displaystyle\lim_{x\to +\infty}x^2(\arctan x - \frac{\pi}{2})+x$

  3. $\displaystyle\lim_{x\to+\infty}\left( \sqrt[3]{x^3+x^2+x+1}-\sqrt{x^2+x+1}\cdot\frac{\ln(e^x+x)}{x} \right)$

  4. $\displaystyle\lim_{x\to 0} \left(\frac{a^x-x\ln a}{b^x-x\ln b} \right)^{\frac{1}{x^2}}$

and I don't know how to touch them. I know: L'Hôpital's rule, Mean value theorem, Taylor's theorem but still don't have this skill. Can anybody help me?

share|cite|improve this question
You'll find a lot of examples and URLs for other examples in the following sci.math thread: – Dave L. Renfro Mar 26 '12 at 19:51
up vote 3 down vote accepted

Usually series expansions are the best way to go. L'Hospital should be avoided whenever possible.

For example in 1)

$$ (x+1)^{1+1/x} = \exp\left((1+\frac1x) \ln(x+1)\right) = \exp\left((1+\frac1x)(\ln(x)+\ln(1+\frac1x))\right)$$ $$ = \exp\left(\ln(x) + \frac{\ln(x)}{x} + \frac1x + O(\frac{1}{x^2})\right) = x \left(1 + \frac{\ln(x)}{x} + \frac{1}{x} + O(\frac{\ln(x)^2}{x^2})\right)$$ $$ = x + \ln(x) + 1 + o(1)$$


$$ x^{1+1/(x+a)} = x \exp\left(\frac{\ln(x)}{x+a}\right) = x \exp\left(\frac{\ln(x)}{x} + O(\frac{\ln (x)}{x^2})\right)$$ $$ = x \left(1 + \frac{\ln(x)}{x} + O(\frac{\ln(x)^2}{x^2})\right)$$ $$ = x + \ln(x) + o(1)$$

so the limit of their difference is $1$.

EDIT: for 2) note that $$\arctan(t) = \frac{\pi}{2} - \arctan(1/t) = \frac{\pi}{2} - \frac{1}{t} + O(\frac{1}{t^3})$$

share|cite|improve this answer
Just out of curiosity, why should we avoid L'Hospital's Rule? – Patrick Mar 21 '12 at 22:20
Series expansions are hardly the best way to go if they’re not in one’s toolbox. – Brian M. Scott Mar 21 '12 at 23:12
The remedy for that is to put them into the toolbox. – Robert Israel Mar 25 '12 at 16:55
just saying, it's de l'Hopital, and NOT de l'Hospital. The internet is full of that typo. – DRC Dec 2 '13 at 18:55
Not that again... See… – Robert Israel Dec 3 '13 at 3:04

(2) Rewrite it as $$\lim_{x\to\infty}\frac{x\arctan x-\frac{\pi}2 x+1}{\frac1x}$$ and apply l’Hospital’s rule twice, followed by a little algebraic simplification.

(4) This appears to succumb to the usual technique for such problems: let $L$ be the desired limit, take logs to get

$$\ln L=\lim_{x\to 0}\frac{\ln(a^x-x\ln a)-\ln(b^x-x\ln b)}{x^2}\;,$$

and beat it to death with l’Hospital’s rule.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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