### Video Transcript

Determine the limit as π approaches β of one minus two over π all raised to the power of eight ππ₯.

In this question, weβre asked to determine a limit. We can see our limit is as π is approaching β. And the first thing we can always try when weβre asked to evaluate a limit is to do this directly. However, if we were to try and do this, we would immediately run into a problem. Inside our parentheses, as π is approaching β, negative two divided by π is approaching zero because the numerator of this fraction remains constant. However, the denominator is growing without bound. And if we look at our exponent, we can see itβs eight multiplied by ππ₯. As π is approaching β, this is going to be approaching β.

So as π approaches β, inside our parentheses, we have one minus zero, so this is approaching one and our exponent is approaching β. So this limit is approaching one to the power of β. And this is an indeterminate form, so we canβt evaluate this limit directly. Weβre going to need to use some other method.

And at this point, we have a lot of different things we could try to evaluate our limit. For example, we could use algebraic manipulation to change the expression inside our parentheses. However, the easiest way to evaluate this limit will be to use one of our limit results involving Eulerβs number. We recall the limit as π’ approaches zero of one plus π’ all raised to the power of one over π’ is equal to Eulerβs constant π. And in fact, we know there are two limit results involving Eulerβs number π. In fact, we can use either of these to evaluate this limit. And itβs very difficult to determine which limit result we should use just by looking at our limit. So if we struggle by using one, we should try switching. So weβre going to try and manipulate our limit so that itβs in the form where we can use our limit result.

To do this, weβre first going to want the expression inside of our parentheses to be in the form one plus π’, and weβll do this by using a substitution. Weβre just going to substitute π’ is equal to negative two over π. So weβve now rewritten our limit as the limit as π approaches β of one plus π’ all raised to the power of eight ππ₯. But this doesnβt help us evaluate our limit yet because now we have a limit as π approaches β, we have π’ is a function in π and π is still in our limit. So weβre going to want to rewrite our entire limit in terms of π’. And to do this, weβre going to need to find an expression for π in terms of π’. And to do this, weβre going to need to rearrange our substitution.

We multiply through by π and then divide through by π’ to get that π is equal to negative two divided by π’. So we now have an expression for π and we can substitute this directly into our limit. However, we also need to know what happens to our value of π’ as π is approaching β. And we can do this by using either of the two equations we got from our substitution. However, itβs easier to see this in the top equation. As π approaches β, negative two over π is approaching zero from the left. This means π’ is approaching zero from the left.

Weβre now ready to use this substitution to rewrite our limit. First, weβve shown as π approaches β, π’ is approaching zero from the left, and this is not strictly necessary. We can just rewrite this as the limit as π’ approaches zero. Since we know if the limit as π’ approaches zero is equal to some number, then the left-hand and right-hand limit as π’ approaches zero will also be equal to that number. Next, we substitute π is equal to negative two over π’ to get the limit as π’ approaches zero of one plus π’ all raised to the power of eight multiplied by negative two over π’ times π₯. Weβre now almost ready to use our limit result. We can see that our limit is as π’ is approaching zero. We can see we have one plus π’ inside of our parentheses. All we need now is to rewrite this in terms of the exponent being one over π’.

So to do this, weβre going to need to simplify our exponent. Since we want an exponent of one over π’, weβll take out a factor of one over π’ in our exponent and then simplify the remaining result. We get our exponent is one over π’ times negative 16π₯. Now, our limit is almost in a form where we can directly use our limit result. Weβre going to want to remove the negative 16π₯ in our exponent from our limit. And since this is an exponent inside of a limit, weβre going to need to use our laws of exponents and the power rule for limits.

First, by using our laws of exponents, we know π to the power of π times π is equal to π to the power of π all raised to the power of π. So inside of our limit, we can rewrite this as one plus π’ all raised to the power of one over π’ all raised to the power of negative 16π₯. And we might want to just directly apply our limit result at this point. However, we canβt do this yet because, remember, the exponent of negative 16π₯ is still inside of our limit. So to take the exponent of negative 16π₯ outside of our limit, weβre going to need to use the power rule for limits. And this will then allow us to use our limit result to evaluate this limit.

Recall the power rule for limits tells us the limit as π approaches π of π of π raised to the πth power is equal to the limit as π approaches π of π of π all raised to the πth power. And we can guarantee that this is true provided the limit as π approaches π of π of π exists and raising this to the πth power exists. And in the case weβre using it, weβll be able to show that both of these two statements are true. And the easiest way to see that both of these prerequisites are true will be to write the next line in our working assuming that our result will hold. By applying the power rule for limits, we would get that this is equal to the limit as π’ approaches zero of one plus π’ all raised to the power of one over π’. And then outside of our limit, we raise all of this to the power of negative 16π₯.

The two prerequisites we need to hold is that the limit inside of this expression exists. Well, we know this is our limit result; we know this is equal to π. We then also need that raising this to our exponent will exist. For example, weβre not allowed to take the square root of a negative number. And of course, in this case, we can see this is true because π is positive. We can always raise this to any exponent. So both of our prerequisites are true, and weβre allowed to use the power rule for limits in this case. Then, all we need to do is use our limit result involving Eulerβs constant π to rewrite this as π to the power of negative 16π₯, which is our final answer.

Therefore, we were able to show the limit as π approaches β of one minus two over π all raised to the power of eight ππ₯ is equal to π to the power of negative 16π₯ for any value of π₯.