# Limit of $(\cos{xe^x} - \ln(1-x) -x)^{\frac{1}{x^3}}$

So I had the task to evaluate this limit

$$\lim_{x \to 0} (\cos{(xe^x)} - \ln(1-x) -x)^{\frac{1}{x^3}}$$

I tried transforming it to:

$$e^{\lim_{x \to 0} \frac{ \ln{(\cos{xe^x} - \ln(1-x) -x)}}{x^3}}$$

So I could use L'hospital's rule, but this would just be impossible to evaluate without a mistake. Also, I just noticed this expression is not of form $\frac{0}{0}$.

Any solution is good ( I would like to avoid Taylor series but if that's the only way then that's okay).

I had this task on a test today and I failed to do it.

• Is it $(\cos x)e^x$ or $\cos (xe^x)$? – Thomas Andrews Feb 12 '15 at 21:55
• The second one, I guess. – Aaron Maroja Feb 12 '15 at 21:56
• It's $\cos{(xe^x)}$ Sorry. – Transcendental Feb 12 '15 at 22:00

First notice that $$\cos (xe^x) = \sum_{n=0}^{\infty}(-1)^n \frac{(xe^x)^{2n}}{2n!}$$

and $$\ln (1 - x) = -\sum_{n=0}^{\infty} \frac{x^n}{n}$$

Thus $$\cos (xe^x) - \ln (1 - x) = \sum_{n=0}^{\infty}(-1)^n \frac{(xe^x)^{2n}}{2n!} +\sum_{n=0}^{\infty}\frac{x^n}{n} = 1 + x - \frac{2x^3}{3} - O(x^4)$$

Therefore we have

$$\ln (1 - \frac{2x^3}{3}) = - \frac{2x^3}{3} - \frac{2x^6}{9} - O(x^9)$$

Finally

\begin{align}\lim_{x \to 0} \frac{\ln (\cos xe^x - \ln (1 - x) - x)}{x^3} &= \lim_{x \to 0} -\frac{2}{3} - \frac{2x^3}{9} - O(x^6) =\color{red}{ -\frac{2}{3}}\end{align}

Now you may find your limit.

• You don't need to write approximately all your statements involving big O notation are correct. – Ethan Feb 12 '15 at 22:47
• @Ethan Thanks!.. – Aaron Maroja Feb 12 '15 at 22:48