# Question About Finding Limits with L'Hopital's Rule

## Homework Statement

A question we had for homework was: Which of the following is equivalent to limh->0 [arcsin((3(x+h))/4) – arcsin (3x/4)]/h ?

## The Attempt at a Solution

We've already walked through the problem in class so I know the steps to solve it, but there's one step that I didn't understand and he couldn't explain quite clearly to me. Why are we allowed to just calculate the derivative of arcsin(3x/4)? I know that according to L'Hopital's Rule we can calculate the limit by finding the limit of the derivatives, but why can we skip the other parts of the function altogether?

It is not an application of L'Hopital's rule. The limit you are given is the definition of the derivative of f(u) = arcsin(3u/4) at the point x. Recall that the derivative of f(x) at the point a is equal to the limit of the difference quotient:
$$f'(a) = \lim_{h\rightarrow 0} \frac{f(a + h) - f(a)}{h}$$
If you see a limit that is of the form on the right side of this equation, and you know the derivative f'(a) by another method, then there is no need to go through the rigmarole of algebraic limit deconstruction: just use your knowledge that it is equal to f'(a).

So you would just be plugging 0 in for h and solving the limit normally?

We cannot replace h by 0 directly, as division by 0 is undefined, and in any case would only be equal to the limit if the expression is continuous at h = 0. Not all functions are continuous or have continuous derivatives!
While it may be possible to simplify the limit by going through extended analytical means, the fact that the limit is equal to the derivative, and we already know the derivative by simpler methods, means we can go ahead and replace the entire limit by its equivalent: the derivative. You should review the definition of the derivative (not just the shortcuts for its calculation) and practice recognizing this type of limit.

I'll definitely do that. Thank you!