I have the following limit to compute:
$$\lim_{x\to 0}{\left(\cos x -1\over {5 x^2}\right)}$$
I need to find the limit as $x \to 0$. I tried using L'Hospital rule , so I found the derivative of the numerator, which is:
$${-\sin x}$$
The derivative of the denominator is: $${10x}$$
Now I have the following:
$${-\sin x \over {10x}}$$
What should I do now? , I'm not sure on how I could apply direct substitution to this problem?
Best Answer
Simply apply L'Hospital's rule again: $$\lim_{x→0}\frac{(-\sin x)'}{(10x)'}$$ $$= \lim_{x→0}\frac{-\cos(x)}{10}$$ $$= \frac{-\cos(0)}{10}$$ $$=-\frac{1}{10}$$