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

If the function 𝑓(π‘₯) = βˆ’3π‘₯⁹ βˆ’ 5, find lim_(β„Žβ†’0) (𝑓(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯))/β„Ž.

02:53

Video Transcript

If the function 𝑓 of π‘₯ equals negative three π‘₯ to the power of nine minus five, find the limit as β„Ž tends to zero of 𝑓 of π‘₯ plus β„Ž minus 𝑓 of π‘₯ over β„Ž.

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

And so substituting this and also our definition for 𝑓 of π‘₯, we get negative three π‘₯ plus β„Ž to the power of nine minus five minus in brackets negative three π‘₯ to the power of nine minus five all divided by β„Ž. 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 we’re ready to worry about the limit. The limit as β„Ž tends to zero of the left-hand side, which is 𝑓 of π‘₯ plus β„Ž minus 𝑓 of π‘₯ over β„Ž, is equal to the limit as β„Ž tends to zero of minus three times π‘₯ plus β„Ž to the power of nine plus three π‘₯ to the power of nine over β„Ž. We can see that’s just substituting directly β„Ž 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 π‘₯ plus β„Ž to the power of nine to get this monstrosity. And we can notice that the terms independent of β„Ž will cancel. We’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„Ž, which we can now take outside the brackets. We now move the common factor of β„Ž outside the brackets, reducing the power of β„Ž in each term by one. And having done that, we can cancel the β„Ž in the numerator with the denominator β„Ž.

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

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