# $\lim_{x \to 0} \frac{e^{\sin^2 x} - \cos x}{x^3}$

$$\lim_{x \to 0} \frac{e^{\sin^2 x} - \cos x}{x^3}$$

Attempt

This is indeterminate of the form $\frac{0}{0}$. Applying L'Hopital's rule twice results in,

$$\lim_{x \to 0} \frac{e^{\sin^2 x}(\sin^2(2x) + \sin^2(4x)) + \cos x}{6x}.$$

Wolfram Alpha says the two sided limit does not exist.

Question

Can I deduce the result by splitting it up into,

$$\frac{1}{6}\lim_{x \to 0} \left(e^{\sin^2 x}(\sin^2(2x) + \sin^2(4x)) + \cos x \right)\left(\frac{1}{x}\right).$$

And saying that the limit does not exist because $\frac{1}{x}$ does not have a two sided limit?

• I suppose. Maybe specify $x\to0^+$ is $+\infty$, $x\to0^-$ is $-\infty$ hence $f$ does not have a two sided limit at 0. Not mandatory though. – user39572 Oct 2 '12 at 12:05
• @JulienGodawatta - Possibly weird qns: I know I can't say the same for $\frac{sin x}{x}$ (which tends to 1 as $x$ tends to 0), but why not? I mean, what is the "algorithm" for working out such things? – Legendre Oct 2 '12 at 12:08
• It matters whether it is of the form $1/0$, or $0/0$.. – Berci Oct 2 '12 at 12:09
• @Legendre L'Hopital's works well in this case (since it is a 0/0) - reduces it to cos! – user39572 Oct 2 '12 at 12:11
• @JulienGodawatta - Yes I know. Just wondering if I can "algorithmize" the thinking this way: 1) if $\frac{0}{0}$ then L'Hopital, 2) if $\frac{a}{0}$, $a$ constant, then split into fractions and check if two sided limits exists for the parts. I suppose I can. Thanks. – Legendre Oct 2 '12 at 12:13

## 3 Answers

$\lim_{x\to 0}\frac{e^x-1}{x}=1\implies e^x\approx 1+x$when $x\to 0$,thus, $e^{\sin^2x}\approx 1+\sin^2x$ as $x\to 0$

Hence, $$\lim_{x \to 0} \frac{e^{\sin^2 x} - \cos x}{x^3}=\lim_{x \to 0} \frac{1+ \sin^2 x - \cos x}{x^3}$$

Now you can easily apply L'Hopital's rule to conclude that limit doesn't exist.

• Nice answer! Is it always "legit" to truncate the Taylor series like this for evaluating limits when $x \to 0$? – Legendre Oct 2 '12 at 12:17
• More technically , one should use $e^x=1+x+o(x)$ – Aang Oct 2 '12 at 12:21
• Yes, and thanks. +1 – Legendre Oct 2 '12 at 12:27

If your calculation is good, then yes. You don't even need to split it into two terms, the fraction already tells you that, when $x\to \pm 0$, the numerator $\to 1$ but the denominator $\to \pm 0$, so it is of the form $\displaystyle\frac1{\pm0}$..

The limit is $\infty$. You can see this in your second limit, as you did, by using L'Hopital.

• Wolfram Alpha says limit does not exist? wolframalpha.com/input/… – Legendre Oct 2 '12 at 12:10
• Yes it is right. I evaluate only by the right side. By the left side it is $\infty$, – Tomás Oct 2 '12 at 13:40