L'Hospital rule to find limit on indeterminate form.

Question

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?

Answer 1

Simply apply L'Hospital's rule again: $$\lim_{x→0}\frac{(-\sin x)'}{(10x)'}$$ $$= \lim_{x→0}\frac{-\cos(x)}{10}$$ $$= \frac{-\cos(0)}{10}$$ $$=-\frac{1}{10}$$

Answer 2

Let me point that the limit as $x\to 0$ of $\:\dfrac{\sin x}x\:$ is a basic high-school result. Therefore, you can write $$\frac{-\sin x}{10 x}=-\frac1{10}\underbrace{\frac{\sin x}x}_{\substack{\downarrow\\1}}.$$