Limit question below!

Question

mortonsalt · Answer

Find a function f(x) and a number a such that:

$$f'(a) = \lim_{x \rightarrow 0}\frac{cosx-1}{x}$$

mortonsalt · Answer

Hello!

anonymous · Answer

Hint: definition of derivative at a point is...

mortonsalt · Answer

Should I try when f(x) = cos x and a = 0?

mortonsalt · Answer

Then when I plug it into the definition of the derivative, I got a=0.

mortonsalt · Answer

But I was wondering if there was another way to solve it.

zenmo · Answer

do you know how to use L'Hospital rule?

mortonsalt · Answer

I was thinking that too. 

So when I use L'Hospital's I get the limit as x approaches 0 is 0.

zenmo · Answer

Then the answer should be 0.

mortonsalt · Answer

Mmm, just double confirming! Thanks guys.

anonymous · Answer

When you get 0/0 you may apply L'hopital's rule, so notice if we plug in 0 we get 0/0 since cos(0) = 1

zenmo · Answer

as long the denominator isn't divided by 0 after applying L'Hospital rule, but in this case, since it's divided by 1. The answer is 0.

anonymous · Answer

$$\lim_{x \rightarrow 0} \frac{ cosx-1 }{ x } = \lim_{x \rightarrow 0} -sinx$$ taking the derivative we get that

mortonsalt · Answer

Yup! Thank you! @iambatman @Zenmo

anonymous · Answer

From the definition,
$$f'(a):=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$Clearly, $a=0$ and $f(x)=\cos x$.

Then $(\cos x)'=-\sin x$, and so $f'(0)=-\sin0=0$.