# How to solve $\lim\limits_{x\to 0} \frac{x - \tan(x)}{x^2}$ Without L'Hospital's Rule?

How to solve $\lim\limits_{x\to 0} \frac{x - \tan(x)}{x^2}$ Without L'Hospital's Rule? you can use trigonometric identities and inequalities, but you can't use series or more advanced stuff.

Note that,

$\frac{\sin(x)-\tan(x)}{x^2} \leq \frac{x-\tan(x)}{x^2}\leq \frac{\tan(x)-\tan(x)}{x^2}=0$, for $\pi/2>x>0$ as $\sin(x) \leq x \leq \tan(x)$ for this interval.

For $x<0$ the inequalities are reversed because the functions tan and sin are even functions $\frac{\sin(x)-\tan(x)}{x^2} \geq \frac{x-\tan(x)}{x^2}\geq \frac{\tan(x)-\tan(x)}{x^2}=0$, for $-\pi/2<x<0$ as $\sin(x) \geq x \geq \tan(x)$ for this interval.

But, $\lim_{x \to 0} \frac{\sin(x)-\tan(x)}{x^2}=\lim_{x \to 0} \frac{\sin(x)}{x} \frac{\cos(x)-1}{x \cos(x)}=\lim_{x \to 0} \frac{-sin^2(x/2)}{(x/2)\cos(x)}=0$

So, $0 \leq \lim_{x \to 0} \frac{x-\tan(x)}{x^2} \leq 0$

Therefore, $\lim_{x \to 0} \frac{x-\tan(x)}{x^2} = 0$

• You can't apply $\sin x\sim x$ in the second equality. – Martín-Blas Pérez Pinilla Jan 12 '15 at 8:36
• @Martin Thanks for pointing out i am changing the answer – Curious Jan 12 '15 at 9:07