Answer
$-16 $
Work Step by Step
Consider: $\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}\dfrac{8x^2}{\cos x-1}$
Need to check that the limit has an indeterminate form.
Thus, $f(0)=\dfrac{8(0)}{\cos 0 -1}=\dfrac{0}{0}$
Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$
This implies:
$\lim\limits_{x \to 0}\dfrac{16x}{-\sin x}=\dfrac{0}{0} $
Again apply L-Hospital's rule .
we get $\lim\limits_{x \to 0}\dfrac{16}{-\cos x}=\dfrac{16}{-\cos 0}=-16 $
