Can someone explain this trigonometric limit without L'Hopital? I can not solve this limit:
$$\lim \limits_{x\to 0}\frac{x^2}{1-\sec(x)}$$
$$\lim \limits_{x\to 0}  \frac{x^2}{1-\sec(x)}=\lim \limits_{x\to 0}\frac {x^2}{1-\sec(x)}\cdot{\frac{1+\sec(x)}{1+\sec(x)}}=\lim \limits_{x\to 0}\frac{x^2(1+\sec(x))}{1-\sec^2(x)}=\lim \limits_{x\to 0}\frac{x^2(1+\sec(x))}{-\tan^2(x)}$$

$\lim \limits_{x\to 0}  \frac{x^2}{1-\sec x}= \lim \limits_{x\to 0} \frac {x^2}{1-\sec x}\cdot{\frac{1+\sec(x)}{1+\sec(x)}}=\lim \limits_{x\to 0} \frac{x^2(1+\sec(x))\cos^2 x}{\cos^2 x-1}=\lim \limits_{x\to 0} \frac{x^2(1+\sec(x))\cos^2 x}{-\sin^2(x)}$
Now if you know the limit of $\frac {\sin x}x$ the other terms are well behaved.
