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

If the function π(π₯) = β3π₯βΉ β 5, find lim_(ββ0) (π(π₯ + β) β π(π₯))/β.

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### Video Transcript

If the function π of π₯ equals negative three π₯ to the power of nine minus five, find the limit as β tends to zero of π of π₯ plus β minus π of π₯ over β.

Before we start worrying about finding the limit, letβs just consider the expression π of π₯ plus β minus π of π₯ over β and in particular this π of π₯ plus β. We have that π of π₯ is negative three π₯ to the power of nine minus five and π of π₯ plus β is just this, but with π₯ replaced by π₯ plus β. So π of π₯ plus β is negative three times π₯ plus β to the power of nine minus five.

And so substituting this and also our definition for π of π₯, we get negative three π₯ plus β to the power of nine minus five minus in brackets negative three π₯ to the power of nine minus five all divided by β. And we can get rid of one the brackets in the numerator; minus negative three π₯ to power of nine minus five is just plus three π₯ to the power of nine plus five. And then we notice that the minus five and the plus five cancel.

And now, I think weβre ready to worry about the limit. The limit as β tends to zero of the left-hand side, which is π of π₯ plus β minus π of π₯ over β, is equal to the limit as β tends to zero of minus three times π₯ plus β to the power of nine plus three π₯ to the power of nine over β. We can see thatβs just substituting directly β equals zero into this right-hand side would give us an indeterminate form. So weβre going to have to be slightly more clever and rewrite this term.

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

And tidying up, we just get this polynomial in π₯ and β, which we can now just directly substitute in to to find the limit as β tends to zero. And this is relatively easy. Thereβs only one term, which doesnβt involve β, negative 27 times π₯ to the power of eight. All the other terms have a factor of β and so after substitution will become zero. And so the limit is simply negative 27π₯ to the power of eight.