# Video: Finding a Limit of a Function Which Can Be Interpreted as the Differential of Another Function

Find lim_(β β 0) (((π₯ + β)β· cos ((π₯ + β)Β³) β (π₯β· cos (π₯Β³)))/β)

03:16

### Video Transcript

Find the limit as β approaches zero of π₯ plus β to the seventh power times cos of π₯ plus β cubed minus π₯ to the seventh power times cos of π₯ cubed all over β.

Now, what weβre not going to do here is to find a way to manipulate the expression in our limit. Instead, weβre going to recognize what this limit actually represents. Remember, the derivative of a function π at a given point π denoted by π prime of π is the limit as β approaches zero of a π of π plus β minus πof π all over β. Now more generally, if we replace π with π₯, we find that the derivative π prime of π₯ is equal to the limit as β approaches zero of π of π₯ plus β minus π of π₯ all over β.

Comparing the general form of the derivative to our limit and we see that π of π₯ must be equal to this function. π of π₯ is equal to π₯ to the seventh power times cos of π₯ cubed. And we know that our limit then is equal to the derivative of this function π prime of π₯. So rather than evaluating the limit, letβs evaluate the derivative. Now, this function π₯ to the seventh power times cos of π₯ cubed is itself the product of two differentiable functions. And so weβre going to use the product rule to find its derivative. This says that the derivative of the product to do differentiable functions is equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯.

Letβs say that π’ is equal to π₯ to the seventh power and π£ is equal to cos of π₯ cubed. According to the product rule, weβre going to need to differentiate each of these with respect to π₯. Well, the derivative of π₯ to the seventh power is quite straightforward. We multiply by the entire exponent and then reduce that exponent by one. So dπ’ by dπ₯, the derivative of π’ with respect to π₯, is seven π₯ to the sixth power. dπ£ by dπ₯ is a little bit more tricky since we have a composite function. And so weβre going to use the chain rule. This says that if π¦ is some function in π’ and π’ itself is some function in π₯, then the derivative of π¦ with respect to π₯ is dπ¦ by dπ’ times dπ’ by dπ₯.

Now, in this case, π¦ is the function in π’. So letβs call that cos of π’, where π’ is π₯ cubed. The derivative of π¦ with respect to π’, the derivative of cos of π’ with respect to π’, is negative sin π’. And then the derivative of π’ with respect to π₯ is three π₯ squared. Replacing π’ with π₯ cubed and we find that dπ£ by dπ₯ is negative three π₯ squared sin of π₯ cubed. Then the derivative of π’ times π£ β in other words, π prime of π₯ β is π’ times dπ£ by dπ₯. So thatβs π₯ to the seventh power times negative three π₯ squared times sin π₯ cubed plus π£ times dπ’ by dπ₯. Thatβs cos of π₯ cubed times seven π₯ to the sixth power.

Letβs rewrite this as seven π₯ to the sixth power times cos of π₯ cubed minus three π₯ to the ninth power times sin of π₯ cubed. Then we can take out a constant factor of π₯ of the sixth power. And this means that π prime of π₯ is π₯ to the sixth power times seven cos of π₯ cubed minus three π₯ cubed times sin of π₯ cubed.

Now, if youβre wondering where the three π₯ cubed comes from, remember, π₯ to the ninth power divided by π₯ to the sixth power is π₯ cubed since we subtract the exponents. And therefore, three π₯ to the ninth power divided by π₯ to the sixth power is three π₯ cubed. And so whilst weβve evaluated π prime of π₯, we also know that this is equal to the limit as β approaches zero of π₯ plus β to the seventh power times cos of π₯ plus β cubed minus π₯ to the seventh power times cos of π₯ cubed over β. Itβs π₯ to the sixth power times seven cos of π₯ cubed minus three π₯ cubed time sin of π₯ cubed.