Video: Evaluating a Given Limit Using L’HΓ΄pital’s Rule

Evaluate lim_(π‘₯ β†’ ∞) (π‘₯Β²/𝑒^(π‘₯)) using L’HΓ΄pital’s rule.

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Video Transcript

Evaluate the limit of π‘₯ squared over 𝑒 to the π‘₯ as π‘₯ approaches ∞ using L’HΓ΄pital’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 ∞ over the limit of 𝑒 to the π‘₯ of the π‘₯ and π‘₯ approaches ∞. The problem is that the limit of π‘₯ squared as π‘₯ approaches ∞ is ∞ as is the limit of 𝑒 to the π‘₯ as π‘₯ approaches ∞. And so we get the indeterminate form ∞ over ∞. This is why we need to use L’HΓ΄pital’s rule. L’HΓ΄pital’s rule says that if the quotient of limits, the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž over the limit of 𝑔 of π‘₯ as π‘₯ approaches π‘Ž, gives an indeterminate form. That’s zero over zero or a positive or negative ∞ over a positive or negative ∞. Then the limit of the quotient of functions 𝑓 of π‘₯ over 𝑔 of π‘₯ as π‘₯ approaches π‘Ž is equal the limit of the quotient of their derivatives 𝑓 prime of π‘₯ over 𝑔 prime of π‘₯ as π‘₯ approaches π‘Ž.

Of course, this only works if 𝑓 and 𝑔 differentiable functions. But in our case, they are 𝑓 of π‘₯ is π‘₯ squared, which is differentiable, as of 𝑔 of π‘₯, which is 𝑒 to the π‘₯. With this choice of 𝑓 of π‘₯ and 𝑔 of π‘₯, we’ve already seen that the quotient to their limits gives the indeterminate form ∞ over ∞. So L’HΓ΄pital’s rule does apply. And hence, we can say that the limit of the quotient of functions that we’re looking for is equal to the limit of the quotient of their derivatives. 𝑓 of π‘₯ is π‘₯ squared, and so 𝑓 prime of π‘₯ is two π‘₯, its derivative. 𝑔 of π‘₯ is 𝑒 to the π‘₯, and its derivative is also 𝑒 to the π‘₯. The exponential function 𝑒 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 ∞. Well, as π‘₯ approaches ∞, two π‘₯ does as well. And the same is true in the denominator. The limit of 𝑒 to the π‘₯ as π‘₯ approaches ∞, as we’ve really seen, is also ∞. We have another indeterminate form here, ∞ over ∞. You might think that we haven’t actually made any progress by applying the L’HΓ΄pital’s rule. But in fact, we have. And we can see this by applying the L’HΓ΄pital’s rule one more time. This time we’re applying it to the limit of two π‘₯ over 𝑒 to the π‘₯ as π‘₯ approaches ∞. And hence, we take 𝑓 of π‘₯ to be two π‘₯ and 𝑔 of π‘₯ to be 𝑒 to the π‘₯.

The derivative of two π‘₯ is two and the derivative of 𝑒 to the π‘₯ is 𝑒 to the π‘₯. And so applying L’HΓ΄pital’s rule a second time, we get the limit of two over 𝑒 to the π‘₯ as π‘₯ approaches ∞. 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 ∞ is just two, and the limit of 𝑒 to the π‘₯ as π‘₯ approaches ∞ is ∞. Two over ∞ isn’t an indeterminate form, although it contains ∞, which isn’t itself a real number. In the context of limits, we can see that any real number over ∞ is zero. And so that’s the value of our initial limit, the limit of π‘₯ squared over 𝑒 to the π‘₯ as π‘₯ approaches ∞ as well.

If you’re worried about saying that two divided by ∞ is zero, then there’s another way of saying this. Instead of finding a quotient of limits, we can think of the function two over 𝑒 to the π‘₯ or two 𝑒 to the negative π‘₯. You might recognize this as an exponential decay, 𝑒 to the negative π‘₯. And hence, two times 𝑒 to the negative π‘₯ approaches zero as π‘₯ approaches ∞. Either way, the answer is zero.

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