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I have the following function, which I am supposed to evaluate:

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

My though is to replace sin x by its Maclaurin polynomial, as such:

$\lim_{x \to 0}{\frac{(e^{-x^2}-1)(x+ O(x^3))}{x \ln (1+x^2)}}$

From here I think I should be able to simplify the denominator, divide everything by x, derivate according to l'Hopital's rule, and get the final result of -1. I'm just not quite sure if this is allowed.

First of all, what happens when I multiply $e^{-x^2}$ by $O(x^3)$ in the denominator? And what happens when I derivate $O(x^3)$? Is there a better strategy?

2 Answers2

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Hint: $$\frac{(e^{-x^2}-1)\sin x}{x \ln (1+x^2)} = -\frac{e^{-x^2}-1}{-x^2} \frac{\sin x}{x} \frac{x^2}{\ln (1+x^2)}.$$

and
$$\lim_{t\to 0} \dfrac{e^t-1}{t} = \lim_{t\to 0} \dfrac{\sin t}{t} = \lim_{t\to 0} \dfrac{\ln (1+t)}{t} = 1.$$

njguliyev
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You can use the Taylor series for the other functions as $$ {\frac{(e^{-x^2}-1)(x+ O(x^3))}{x \ln (1+x^2)}}= {\frac{({-x^2}+O(x^4))(x+ O(x^3))}{x (x^2+O(x^4))}}\sim \frac{(-x^2)(x)}{(x)(x^2)}=-1. $$

Mhenni Benghorbal
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  • That looks promising! Is the multiplication of $-x^2$ by $O(x^3) = O(x^5)$, and $O(x^4) O(x^3) = O(x^7)$? Thanks! –  Oct 13 '13 at 16:27
  • @JimmyC: $O(x^3)O(x^4)=O(x^7)$ it is a well known fact a bout the big O notation. For the first one, try to use the following fact $\lim_{x\to x_0}\frac{f(x)}{g(x)}=c \implies f=O(g) $ and see what you get. See here. – Mhenni Benghorbal Oct 13 '13 at 16:40