0
$\begingroup$

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

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?

  • 0
    @Legendre Ah. I suppose. Bear in mind L'Hopital doesn't systematically work though...2012-10-02

3 Answers 3

3

$\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.

  • 0
    Yes, and thanks. +12012-10-02
1

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 $\displaystyle\frac1{\pm0}$..

0

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

  • 0
    Yes it is right. I evaluate only by the right side. By the left side it is $\infty$,2012-10-02