### Video Transcript

Find the limit as π₯ approaches zero of one plus π₯ all cubed minus one all divided by one plus five π₯ to the fifth power minus one.

The question wants us to evaluate the limit of a rational function. Since a rational function is the quotient of polynomials, we can attempt to evaluate this by using direct substitution. Substituting π₯ is equal to zero gives us one plus zero cubed minus one divided by one plus five times zero to the fifth power minus one. And if we evaluate this expression, we see we get the indeterminate form zero divided by zero. So, we canβt evaluate this limit by using direct substitution. So, weβre going to need a different approach to evaluate this limit.

We recall we can attempt to evaluate the limit of the quotient of two functions by using LβHΓ΄pitalβs rule. So, we recall the following version of LβHΓ΄pitalβs rule. If we have two differentiable functions π and π, and the derivative of π is not equal to zero around some value π. Although the derivative of π as π is allowed to be zero. And we know the limit as π₯ approaches π of π of π₯ and the limit as π₯ approaches π of π of π₯ both exist and are equal to zero. Then by this version of LβHΓ΄pitalβs rule, the limit as π₯ approaches π of the quotient of π of π₯ and π of π₯ is equal to the limit as π₯ approaches π of the quotient of π prime of π₯ and π prime of π₯. LβHΓ΄pitalβs rule allows us to evaluate some indeterminate limits.

To apply this version of the LβHΓ΄pitalβs rule to the limit given to us in the question, weβll first set our function π of π₯ to be the function in the numerator of our limit. Thatβs one plus π₯ all cubed minus one. And weβll set π of π₯ to be the function in the denominator of our limit. Thatβs one plus five π₯ all to the fifth power minus one.

Finally, since we want the limit as π₯ approaches zero, weβll set π equal to zero. We now need to check that our prerequisites for LβHΓ΄pitalβs rule are true. First, we need to check that both of our functions π and π are differentiable. We can see that both π of π₯ and π of π₯ are polynomials. And polynomials are differentiable for all real numbers. So, our first prerequisite for LβHΓ΄pitalβs rule is true.

Next, we need to check that π prime of π₯ is not equal to zero around π₯ is equal to zero. Thereβs a few different ways we could differentiate π of π₯. For example, we could expand our parenthesis and then differentiate this by using the power rule for differentiation. However, we could also notice that the first term in our function of π of π₯ is a composite function. So, we could differentiate this by using the chain rule.

If we set some function π’ of π₯ to be one plus five π₯, then we have that π of π₯ is equal to π’ of π₯ to the fifth power minus one. We now see that π is a function of π’ and π’ is a function of π₯. And we recall the chain rule tells us if we have two functions π£ and π€ where π£ is a function of π€ and π€ is a function of π₯, then the derivative of π£ of π€ with respect to π₯ is equal to dπ£ dπ€ times dπ€ dπ₯.

So, applying the chain rule to π of π₯, we have that π prime of π₯ is equal to the derivative of π’ to the fifth power minus one with respect to π’ times dπ’ by dπ₯. And we could evaluate both of these derivatives by using the power rule for differentiation. We see that dπ’ by dπ₯ is equal to the derivative of one plus five π₯ with respect to π₯. The derivative of the constant one is equal to zero. And the derivative of five π₯ is just equal to five.

And to differentiate π’ to the fifth power minus one with respect to π’, we want to multiply it by our exponent and then reduce the exponent by one. So, to differentiate π’ to the fifth power with respect to π’, we get five π’ to the fourth power. And then the derivative of negative one with respect to π’ is just equal to zero. So, weβve shown that π prime of π₯ is equal to 25π’ to the fourth power. Remember that we defined π’ to be one plus five π₯. So, rewriting π’ as one plus five π₯, weβve shown that π prime of π₯ is equal to 25 times one plus five π₯ to the fourth power.

Remember, we want to show that π prime of π₯ is not equal to zero around π₯ is equal to zero. So, letβs consider the values of π₯ where π prime of π₯ could be equal to zero. We see that π prime of π₯ is a product of functions. And a product of functions can only be equal to zero if one of those factors is equal to zero. And of course, 25 is not equal to zero. So, the only way that our function π prime of π₯ can be equal to zero is if one plus five π₯ is equal to zero. And we can rearrange this linear factor to see π₯ is equal to negative one-fifth.

At first, we might be worried this means we canβt use LβHΓ΄pitalβs rule. However, we can just choose an interval around π₯ is equal to zero which does not include π₯ is equal to negative one-fifth. For example, we know π prime of π₯ cannot be equal to zero on the closed interval from negative one-tenth to one-tenth. So, weβve shown that π prime of π₯ is not equal to zero around π₯ is equal to zero.

Next, we need the limit as π₯ approaches zero of π of π₯ and the limit as π₯ approaches zero of π of π₯ to be equal to zero. However, weβve actually shown that this is already the case. When we were initially trying to evaluate our limit by direct substitution, we showed that the limit as π₯ approaches π of π of π₯ is equal to zero and the limit as π₯ approaches zero of π of π₯ was equal to zero. Theyβre just the numerator and the denominator of this limit, respectively.

So, weβve shown that all of our prerequisites for this version of LβHΓ΄pitalβs rule are true. So, now, instead of calculating the limit given to us in the question, we can instead calculate the limit as π₯ approaches zero of π prime of π₯ divided by π prime of π₯. Weβve already found an expression for π prime of π₯, so letβs find an expression for π prime of π₯.

We have that π of π₯ is equal to one plus π₯ all cubed minus one. And we could differentiate this by using the chain rule as we did with π of π₯. However, we could also just expand our parenthesis. We can expand these parentheses by either using the FOIL method or by using binomial expansion. Weβll use binomial expansion. Since three choose one and three choose two are both equal to three, we get one cubed plus three times one times π₯ squared plus three times one squared times π₯ plus π₯ cubed minus one, which simplifies to give us π₯ cubed plus three π₯ squared plus three π₯.

Now, we can just differentiate this by using the power rule for the differentiation. We get π prime of π₯ is equal to three π₯ squared plus six π₯ plus three. So, now that we found expressions for π prime of π₯ and π prime of π₯, we can rewrite the limit as π₯ approaches zero of π prime of π₯ divided by π prime of π₯ as the limit as π₯ approaches zero of three π₯ squared plus six π₯ plus three all divided by 25 times one plus five π₯ to the fourth power.

And we see that this is the limit of a rational function. So, we can attempt to evaluate this by using direct substitution. Substituting π₯ is equal to zero into this quotient gives us three times zero squared plus six time zero plus three all divided by 25 times one plus five times zero to the fourth power. And finally, we can evaluate this expression to get three divided by 25. So, weβve shown, by using LβHΓ΄pitalβs rule, the limit as π₯ approaches zero of one plus π₯ all cubed minus one divided by one plus five π₯ to the fifth power minus one is equal to three divided by 25.