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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.

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  • $\begingroup$ Is it $(\cos x)e^x$ or $\cos (xe^x)$? $\endgroup$ Commented Feb 12, 2015 at 21:55
  • $\begingroup$ The second one, I guess. $\endgroup$ Commented Feb 12, 2015 at 21:56
  • $\begingroup$ It's $\cos{(xe^x)}$ Sorry. $\endgroup$ Commented Feb 12, 2015 at 22:00

1 Answer 1

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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.

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  • $\begingroup$ You don't need to write approximately all your statements involving big O notation are correct. $\endgroup$ Commented Feb 12, 2015 at 22:47
  • $\begingroup$ @Ethan Thanks!.. $\endgroup$ Commented Feb 12, 2015 at 22:48

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