# Calculus Problem of the Week October 28, 2011

Sometimes easy and sometimes hard, our calculus problem of the week could come from any calculus topic. If you really want to get better at calculus, following these problems is a great way to make yourself practice! Past calculus problems of the week.

This week’s problem:
(click “see the solution” at the bottom of post to, well, see the solution. If it doesn’t work, please click the title of this post and then try (I’m working on this!))

Last week’s problem stated: “Let $f(x)$ be a differentiable function at $x=0$ and $\lim_{x->0}\dfrac{f(x)}{x}=4$. Find $f(0)$“.

THIS WEEK: Find $f'(0)$. Hint: This will involve a bit of a “trick” in some sense of the word!

See the solution.

Recall that last week we figured out that $f(0)$ must be zero. Using that, let’s take a look at another way to write the original limit: $\lim_{x->0}\dfrac{f(x)}{x}=\lim_{x->0}\dfrac{f(x)-f(0)}{x-0}$

Since we know $f(0)=0$ we can rewrite it like this. The question is WHY would we do that? Look really closely. Does that look familiar? What if I changed the x to h? This limit is actually the limit definition of $f'(0)$! That means that $f'(0)=4$ since the limit is 4. A little tricky, but it was that step of realizing how to use the results from last week that got us going!