# Video: Finding Limits Using the Definition of a Derivative as a Limit

If the function 𝑓(𝑥) = −3𝑥⁹ − 5, find lim_(ℎ→0) (𝑓(𝑥 + ℎ) − 𝑓(𝑥))/ℎ.

02:53

### Video Transcript

If the function 𝑓 of 𝑥 equals negative three 𝑥 to the power of nine minus five, find the limit as ℎ tends to zero of 𝑓 of 𝑥 plus ℎ minus 𝑓 of 𝑥 over ℎ.

Before we start worrying about finding the limit, let’s just consider the expression 𝑓 of 𝑥 plus ℎ minus 𝑓 of 𝑥 over ℎ and in particular this 𝑓 of 𝑥 plus ℎ. We have that 𝑓 of 𝑥 is negative three 𝑥 to the power of nine minus five and 𝑓 of 𝑥 plus ℎ is just this, but with 𝑥 replaced by 𝑥 plus ℎ. So 𝑓 of 𝑥 plus ℎ is negative three times 𝑥 plus ℎ to the power of nine minus five.

And so substituting this and also our definition for 𝑓 of 𝑥, we get negative three 𝑥 plus ℎ to the power of nine minus five minus in brackets negative three 𝑥 to the power of nine minus five all divided by ℎ. And we can get rid of one the brackets in the numerator; minus negative three 𝑥 to power of nine minus five is just plus three 𝑥 to the power of nine plus five. And then we notice that the minus five and the plus five cancel.

And now, I think we’re ready to worry about the limit. The limit as ℎ tends to zero of the left-hand side, which is 𝑓 of 𝑥 plus ℎ minus 𝑓 of 𝑥 over ℎ, is equal to the limit as ℎ tends to zero of minus three times 𝑥 plus ℎ to the power of nine plus three 𝑥 to the power of nine over ℎ. We can see that’s just substituting directly ℎ equals zero into this right-hand side would give us an indeterminate form. So we’re going to have to be slightly more clever and rewrite this term.

We can binomially expand 𝑥 plus ℎ to the power of nine to get this monstrosity. And we can notice that the terms independent of ℎ will cancel. We’ve got a negative three times 𝑥 to the power of nine and a plus three times 𝑥 to the power of nine. And everything else has a factor of ℎ, which we can now take outside the brackets. We now move the common factor of ℎ outside the brackets, reducing the power of ℎ in each term by one. And having done that, we can cancel the ℎ in the numerator with the denominator ℎ.

And tidying up, we just get this polynomial in 𝑥 and ℎ, which we can now just directly substitute in to to find the limit as ℎ tends to zero. And this is relatively easy. There’s only one term, which doesn’t involve ℎ, negative 27 times 𝑥 to the power of eight. All the other terms have a factor of ℎ and so after substitution will become zero. And so the limit is simply negative 27𝑥 to the power of eight.