Question Video: Solving Limits by Transforming Them into the Natural Exponent Limit Forms | Nagwa Question Video: Solving Limits by Transforming Them into the Natural Exponent Limit Forms | Nagwa

# Question Video: Solving Limits by Transforming Them into the Natural Exponent Limit Forms Mathematics • Third Year of Secondary School

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Determine lim_(π₯ β β)(π₯ + 4/π₯ β 4)^(π₯ β 3)).

07:31

### Video Transcript

Determine the limit as π₯ approaches β of π₯ plus four all divided by π₯ minus four all raised to the power of π₯ minus three.

In this question, weβre asked to evaluate a limit. And the first thing we should do whenever weβre asked to evaluate a limit is trying to do it directly. And thereβs two potential ways we could try and evaluate this limit. Our limit is as π₯ is approaching β. We could check what happens inside of our parentheses. Well, in our numerator as π₯ is approaching β, our numerator is unbounded; itβs growing to β. The same is true in our denominator; itβs also growing to β. And in fact, the same is true in our exponent. π₯ minus three approaches β as π₯ approaches β. So this gives us β over β all raised to the power of β. And this is an indeterminant form because, for example, β over β is an indeterminant form. So this doesnβt help us answer our question.

But there is another way we could have tried to do this directly. Inside of our parentheses, we have a rational function. And we know a few different ways of evaluating the limit as π₯ approaches β of a rational function. For example, we could use polynomial division or look at the coefficients of the leading terms and the powers of the leading terms. And using either of these methods is possible to prove the limit of the rational function inside of the parentheses as π₯ approaches β is equal to one. However, our exponent is still approaching β as π₯ approaches β. So this gives us one to the power of β, which is also an indeterminant form.

So we just canβt evaluate this limit directly. Weβre going to need to try and use a different method. And to do this, letβs start by dividing through the rational function inside of our parentheses. And in fact we do know a few different ways of doing this. For example, we could directly do this by using polynomial division. And this would work; however, there is another method we could use. We can see the leading term in both of our numerator and our denominator have a coefficient of one. And theyβre both of the same power. This means our denominator is going to go into our numerator one time.

Then we know we need to add some constant divided by our denominator π₯ minus four. And if we solve this, we can just show itβs equal to positive eight. Either method gives us the following limit: the limit as π₯ approaches β of one plus eight divided by π₯ minus four all raised to the power of π₯ minus three. And now we can see this limit is very similar to a limit we already know. This looks very similar to our limit involving Eulerβs number π. The limit as π approaches β of one plus one over π all raised to the πth power is equal to π. So weβre going to try and use this limit result to help us evaluate this limit.

Before we do this, itβs also worth pointing out we could have used our other limit result involving Eulerβs number π. Usually, one of the two limit results will give us easier working out than the other one. However, it is very difficult to see which limit result we should use just by looking at the limit we need to evaluate. So if you do get stuck trying to use one of these results, try switching to the other one. Now to use this limit result, weβre going to need to rewrite our limit. We see inside of our parentheses we need one plus one over π.

So to write our limit in this form, weβre going to use the substitution. Weβre going to set one over π to be equal to eight divided by π₯ minus four. So by using the substitution, we can rewrite our limit as the limit as π₯ approaching β of one plus one over π all raised to the power of π₯ minus three. But itβs difficult to evaluate this limit directly because it contains both π₯βs and πβs. So letβs try and rewrite this limit entirely in terms of π. And to do this, weβre going to need to find an expression for π₯ in terms of π. And we can do this by using our substitution. Weβre going to want to rearrange this equation to get an expression for π₯ in terms of π.

To do this, weβll start by taking the reciprocal of both sides of the equation. This gives us π is equal to π₯ minus four all divided by eight. Next, weβll multiply both sides of this equation through by eight. This gives us eight π is equal to π₯ minus four. Finally, we just add four to both sides of the equation. This gives us eight π plus four is equal to π₯. Now weβll substitute this expression for π₯ into our limit. Doing this, we get the limit as π₯ approaches β of one plus one over π all raised to the power of eight π plus four minus three.

And of course, we can simplify this. In our exponent, we have four minus three, which is, of course, just equal to one. So we finished to rewrite our limit in the following form. However, we still need to see what happens to our values of π as π₯ approaches β. And the easiest way to do this is to look at our expression for π₯ in terms of π. As π₯ approaches β, the right-hand side of this equation is growing without bound. This means the left-hand side of this equation must also be growing without bounds. And the only way that this can happen is if our values of π are growing without bound. In other words, as π₯ approaches β, π must be approaching β.

Therefore, by using this, we can rewrite our limit as the limit as π approaches β of one plus one over π all raised to the power of eight π plus one. And now we can see this limit is almost exactly in the form of our limit rule. The only difference is instead of our exponent being π, our exponent is eight π plus one. So weβre going to want to try and rewrite this limit to have the exponent as just π. And weβll do this by using our laws of exponents. First, we can see our exponent is a sum. So weβll rewrite this by using the fact that π to the power of π plus π is equal to π to the power π multiplied by π to the power of π.

So by setting our value of π equal to one plus one over π, π equal to eight π, and π equal to one, we can rewrite our limit as the limit as π approaches β of one plus one over π all raised to the power of eight π multiplied by one plus one over π all raised to the power of one. And of course, raising something to the power of one doesnβt change its value. But now we can see weβre taking the limit of a product. So we can try and evaluate this limit by using the product rule for limits. The product rule for limits tells us the limit of a product is equal to the product of a limit provided the limit of both of our factors exists.

And in fact, when we answer this question, weβll show that both of these limits exist directly. Letβs start with the limit as π approaches β of one plus one over π. We can evaluate this limit directly. As π approaches β, one over π is approaching zero, since its denominator is growing without bound and its numerator remains constant. So as π approaches β, this factor is just approaching one. Therefore, by using the product rule for limits, we can rewrite our limit as the limit as π approaches β of one plus one over π all raised to the power of eight π provided this limit exists.

And now this is almost exactly in the form of our limit result; however, our exponent is eight π instead of π. So weβre going to use one more of our laws of exponents. This time, weβre going to use the fact that π to the power of π multiplied by π is equal to π to the πth power all raised to the power of π. So by using this, we were able to rewrite our limit as limit as π approaches β of one plus one over π all raised to the πth power all raised to the eighth power. But now we can see weβre evaluating the limit of a function raised to the eighth power. So we can try and evaluate this limit by using the power rule for limits.

We recall one version of the power rule for limits tells us the limit as π₯ approaches β of π of π₯ raised to the πth power is equal to the limit as π₯ approaches β of π of π₯ all raised to the πth power. And we can guarantee this is true provided π is an integer and the limit as π₯ approaches β of π of π₯ exists. And in this case, both of these conditions are true. Our value of π is eight, which is an integer. And the limit we need to exist is the limit as π approaches β of one plus one over π all raised to the πth power, which is in fact our limit result. We already know this is equal to π.

Therefore, by using the power rule for limits, instead of raising the function inside of our elements to the eighth power, we can instead just raise our entire limit to the eighth power. And finally, we can just evaluate this limit by using our limit rule. We know itβs equal to π. And then we need to raise this entire expression to the eighth power. And this gives us our final answer of π to the eighth power. Therefore, we were able to show the limit as π₯ approaches β of π₯ plus four all divided by π₯ minus four all raised to the power of π₯ minus three is equal to π to the eighth power.

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