# Question Video: Evaluating a Given Limit Using LβHΓ΄pitalβs Rule Mathematics • Higher Education

Evaluate lim_(π₯ β β) (π₯Β²/π^(π₯)) using LβHΓ΄pitalβs rule.

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### Video Transcript

Evaluate the limit of π₯ squared over π to the π₯ as π₯ approaches β using LβHΓ΄pitalβs rule.

We might try to evaluate this limit by using the fact that the limit of a quotient of functions is a quotient of their limits. So we get the limit of π₯ squared as π₯ approaches β over the limit of π to the π₯ of the π₯ and π₯ approaches β. The problem is that the limit of π₯ squared as π₯ approaches β is β as is the limit of π to the π₯ as π₯ approaches β. And so we get the indeterminate form β over β. This is why we need to use LβHΓ΄pitalβs rule. LβHΓ΄pitalβs rule says that if the quotient of limits, the limit of π of π₯ as π₯ approaches π over the limit of π of π₯ as π₯ approaches π, gives an indeterminate form. Thatβs zero over zero or a positive or negative β over a positive or negative β. Then the limit of the quotient of functions π of π₯ over π of π₯ as π₯ approaches π is equal the limit of the quotient of their derivatives π prime of π₯ over π prime of π₯ as π₯ approaches π.

Of course, this only works if π and π differentiable functions. But in our case, they are π of π₯ is π₯ squared, which is differentiable, as of π of π₯, which is π to the π₯. With this choice of π of π₯ and π of π₯, weβve already seen that the quotient to their limits gives the indeterminate form β over β. So LβHΓ΄pitalβs rule does apply. And hence, we can say that the limit of the quotient of functions that weβre looking for is equal to the limit of the quotient of their derivatives. π of π₯ is π₯ squared, and so π prime of π₯ is two π₯, its derivative. π of π₯ is π to the π₯, and its derivative is also π to the π₯. The exponential function π to the π₯ has the property that its derivative is just itself.

Now, we can try to use the fact that the limit of a quotient of functions is a quotient of their limits. What is the limit of two π₯ as π₯ approaches β. Well, as π₯ approaches β, two π₯ does as well. And the same is true in the denominator. The limit of π to the π₯ as π₯ approaches β, as weβve really seen, is also β. We have another indeterminate form here, β over β. You might think that we havenβt actually made any progress by applying the LβHΓ΄pitalβs rule. But in fact, we have. And we can see this by applying the LβHΓ΄pitalβs rule one more time. This time weβre applying it to the limit of two π₯ over π to the π₯ as π₯ approaches β. And hence, we take π of π₯ to be two π₯ and π of π₯ to be π to the π₯.

The derivative of two π₯ is two and the derivative of π to the π₯ is π to the π₯. And so applying LβHΓ΄pitalβs rule a second time, we get the limit of two over π to the π₯ as π₯ approaches β. Now, we can use the fact that the limit of a quotient is a quotient of the limits. The limit of two as π₯ approaches β is just two, and the limit of π to the π₯ as π₯ approaches β is β. Two over β isnβt an indeterminate form, although it contains β, which isnβt itself a real number. In the context of limits, we can see that any real number over β is zero. And so thatβs the value of our initial limit, the limit of π₯ squared over π to the π₯ as π₯ approaches β as well.

If youβre worried about saying that two divided by β is zero, then thereβs another way of saying this. Instead of finding a quotient of limits, we can think of the function two over π to the π₯ or two π to the negative π₯. You might recognize this as an exponential decay, π to the negative π₯. And hence, two times π to the negative π₯ approaches zero as π₯ approaches β. Either way, the answer is zero.