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I got this exercise which I quite frankly can't wrap my head around

$$\lim_{x\rightarrow 0} \frac{\ln(1-x)+\sin(x)}{x^2 e^x} $$

The result should be:$\ -\frac{1}{2} $

I tried by derivating the whole function and that led me nowhere. If I put in $\ x = 0 $, I get $\ \frac{0}{0} $ which means I can apply L'Hôpital's rule but that led me to the wrong answer as well. Obviously I'm doing something wrong. I had a look at Wolfram Alphas step-by-step solution to this exercise but I didn't really understand it.

I'm appreciative with any possible help.

Thanks,

Michael.

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2 Answers 2

up vote 7 down vote accepted

The $e^x$ part safely has limit $1$, so we can forget about it and calculate $$\lim_{x\to 0}\frac{\ln(1-x)+\sin x}{x^2}.$$ Now L'Hospital's Rule works nicely, with smaller probability of error than if we keep the $e^x$ part.

I prefer a somewhat more "hands on" approach that uses the first few terms of the power series expansions of $\ln(1-x)$ and $\sin x$. Note that $\ln(1-x)=-x-\dfrac{x^2}{2}+o(x^2)$ while $\sin x=x+o(x^2)$. Add. We get $-\dfrac{x^2}{2}+o(x^2)$. Divide by $x^2$. So our expression is $-\dfrac{1}{2}+o(1)$, which is the desired result.

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Using L'Hospital's rule seems like a good way to go.

Taking the derivative of the top and bottom you get $$\frac{ - \frac{1}{1-x} + \cos(x)}{2x e^x + x^2 e^x}$$

This is still $\frac{0}{0}$ so you can use L'Hospital again to get $$ \frac{-\frac{1}{(1-x)^2} - \sin(x)}{2e^x + 2xe^x + 2xe^x + x^2 e^x}$$ which evaluates to $-\frac{1}{2}$

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