# How do you evaluate the limit #lim(5x^2)/(2x)dx# as #x->0#?

##### 1 Answer

#### Explanation:

The first line of attack when looking at a limit should be to plug in the number we want the function to approach.

In this case we get a weird

This is an indeterminate form !

We solve indeterminate forms by using L'Hôpital's Rule.

If you're new to calculus, this might look a little messy, but I'll try to keep it simple, so bear with me. You can even skip the mumbo-jumbo and just read the math part if you don't need the theory.

L'Hôpital's Rule says that if

then

(

Essentially, whichever one gets where it's going faster (i.e. has a greater rate of change) wins.

In our case, the numerator is trying to go to zero and the denominator is trying to go to infinity, so if the numerator is growing faster we'll get

Now onto the math:

We've established that plugging in zeros to

which means, by L'Hôpital's Rule:

**note: There are also many cases in which you do L'Hôpital's Rule and get another indeterminate form, in which case you have to apply L'Hôpital's Rule more than once.**