# Calculate difficult limits

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?

-
You'll find a lot of examples and URLs for other examples in the following sci.math thread: tinyurl.com/ccy4r5d – Dave L. Renfro Mar 26 '12 at 19:51

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

For example in 1)

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

while

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

so the limit of their difference is $1$.

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

-
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 math.stackexchange.com/questions/179680/… – 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.

-