### Video Transcript

Find the limit as π₯ goes to infinity of two π₯ multiplied by π to the negative two π₯.

Firstly, we will use the fact that π₯ to the power of negative π is equal to one over π₯ to the power of π. And we can rewrite our limit as the limit as π₯ goes to infinity of two π₯ over π to the two π₯. Now, letβs label the numerator of our function as π of π₯ and the denominator of our function as π of π₯. Letβs consider the limit of π and π separately as π₯ tends to infinity. The limit as π₯ approaches infinity of π of π₯, which is the same as the limit as π₯ approaches infinity of two π₯, is simply equal to infinity. And the limit as π₯ approaches infinity of π of π₯, which we can write as the limit as π₯ approaches infinity of π to the two π₯, will also be equal to infinity.

The reason why we can say both of these limits are equal to infinity is because as π₯ gets larger and larger and larger, both of our functions π and π are also getting larger, since theyβre both increasing functions. Now that we know the limits of both π and π, we can say that the limit as π₯ approaches infinity of π of π₯ over π of π₯ is equal to infinity over infinity. And this is undefined.

However, since our limit is equal to infinity over infinity, this tells us that we may be able to use LβHΓ΄pitalβs rule. LβHΓ΄pitalβs rule tells us that for functions π and π, both of which are differentiable near π, with the limit as π₯ approaches π of π of π₯ equal to the limit as π₯ approaches π of π of π₯ which is equal to zero or positive or negative infinity, where π is a real number, infinity, or negative infinity, then the limit as π₯ approaches π of π of π₯ over π of π₯ is equal to the limit as π₯ approaches π of π prime of π₯ over π prime of π₯.

Now, weβve already found that the limits as π₯ approaches infinity of both π of π₯ and π of π₯ are equal to positive infinity. And so therefore, this condition of LβHΓ΄pitalβs rule is satisfied. We have that our value of π is positive infinity. Therefore, this condition is also satisfied. Now, all we require is for both π and π to be differentiable. π is equal to two π₯ which is a polynomial. And therefore, it is differentiable. π is equal to π to the two π₯ which is an exponential, which is also differentiable. Therefore, this condition is also satisfied.

Our next step is to find π prime of π₯ and π prime of π₯, where the prime denotes a differential with respect to π₯. We have that π of π₯ is equal to two π₯. Differentiating this with respect to π₯ gives us that π prime of π₯ is equal to two. Next up, we have that π of π₯ is equal to π to the power of two π₯. Differentiating this function with respect to π₯, we find that π prime of π₯ is equal to two π to the two π₯. We are now ready to use LβHΓ΄pitalβs rule. We recall that we can write the limit which weβre trying to find in the form the limit as π₯ approaches infinity of two π₯ over π to the two π₯.

Using LβHΓ΄pitalβs rule, we can write this as the limit as π₯ approaches infinity of two over two π to the two π₯. From here, we can cancel a factor of two in the numerator and denominator, leaving us with the limit as π₯ approaches infinity of one over π to the two π₯. Now, we have already established that the limit as π₯ approaches infinity of π to the two π₯ is equal to infinity. And so we know that as π₯ is getting larger and larger, the denominator of our function, which is π to the two π₯, will also be getting larger and larger. Therefore, one over π to the two π₯ will be getting closer and closer to zero.

And so, we can say that this limit is equal to zero. Therefore, we can conclude that the solution to this problem is that the limit as π₯ approaches infinity of two π₯ π to the negative two π₯ is equal to zero.