### Video Transcript

Evaluate the limit as π₯ approaches zero of one minus the cos of π₯ all divided by π to the power of π₯ minus π₯ minus one.

We see weβre asked to evaluate the limit as π₯ approaches zero of the quotient of two functions. We see our numerator is one minus the cos of π₯. Thatβs a constant minus a standard trigonometric function. And we see our denominator is π to the power of π₯ minus π₯ minus one. Thatβs an exponential function minus a linear function. So, we can evaluate all of these by direct substitution. So, we can attempt to evaluate our limit by direct substitution.

To attempt to evaluate this limit by direct substitution, we substitute π₯ is equal to zero. This gives us one minus the cos of zero all divided by π to the zeroth power minus zero minus one. We know the cos of zero is one and π to the zeroth power is also one. So, this simplifies to give us one minus one divided by one minus one, which is just equal to zero over zero. So, direct substitution left us with the indeterminate form zero over zero. This means weβre going to need to try some other method to evaluate this limit.

We recall LβHΓ΄pitalβs rule gives us another method of evaluating a limit if it gives us an indeterminate form. Weβll use the following version of LβHΓ΄pitalβs rule. If we have two functions π and π, which are differentiable, and π prime of π₯ is not equal to zero around some value of π₯ equal to π. Although π prime of π₯ is allowed to be equal to zero when π₯ is equal to π, where the limit as π₯ approaches π of π of π₯ and the limit as π₯ approaches π of π of π₯ are both equal to zero. Then the limit as π₯ approaches π of π of π₯ divided by π of π₯ is equal to the limit as π₯ approaches π of π prime of π₯ divided by π prime of π₯.

In other words, under these conditions, if weβre calculating the limit as π₯ approaches π and we get the indeterminate form of zero divided by zero, we can instead try to evaluate the limit as π₯ approaches π of π prime of π₯ divided by π prime of π₯. So to use this version of LβHΓ΄pitalβs rule, weβll set our function π of π₯ to be the function in the numerator of our limit β thatβs one minus the cos of π₯ β and our function π of π₯ to be the function in the denominator of our limit β thatβs π to the power of π₯ minus π₯ minus one. And since weβre taking the limit as π₯ approaches zero, weβll set our value of π equal to zero.

The first thing we need to check to use LβHΓ΄pitalβs rule is that both our function π and our function π are differentiable. And itβs actually possible to see this directly from the definitions of our function π and our function π. However, since we need to calculate the limit as π₯ approaches zero of π prime of π₯ divided by π prime of π₯, itβs also useful to find expressions for π prime of π₯ and π prime of π₯. So, letβs start by finding π prime of π₯.

We know the derivative of one with respect to π₯ is zero and the derivative of negative the cos of π₯ is equal to the sin of π₯. Now, letβs find π prime of π₯. We know the derivative of π to the power of π₯ with respect to π₯ is just equal to itself. Next, the derivative of negative π₯ is equal to negative one. Finally, the derivative of the constant negative one is just equal to zero. So, we have that π prime of π₯ is equal to the sin of π₯ and π prime of π₯ is equal to π to the power of π₯ minus one. So, both π and π are differentiable.

Next, we need to show that π prime of π₯ is not equal to zero around our value of π₯ is equal to π, which in this case is zero. And at first, it seems we immediately run into a problem. We see if we set π prime of π₯ equal to zero β so thatβs π to the power of π₯ minus one is equal to zero β we can then solve this equation for π₯. Weβll add one to both sides of our equation. And then either by inspection or by taking the natural logarithms of both sides of this equation, we see that we get π₯ is equal to zero.

So, it seems we have a problem: π prime of zero is equal to zero. But we remember in our version of LβHΓ΄pitalβs rule, π prime of π is allowed to be equal to zero. And π is equal to zero. So, weβre allowed that π prime of zero is equal to zero. So, the only point where π prime of π₯ is equal to zero is when π₯ is equal to zero. So, weβve shown that our second prerequisite for LβHΓ΄pitalβs rule is also true. Next, we need to show the limit as π₯ approaches zero of π of π₯ and the limit as π₯ approaches zero of π of π₯ are both equal to zero. But weβve actually already done this when we attempted to evaluate our limit by using direct substitution.

If we look at the numerator and the denominator separately in this calculation, we can see the numerator is actually just the limit of π of π₯ as π₯ approaches zero and the denominator is the limit as π₯ approaches zero of π of π₯. And we found both of our numerator and our denominator evaluated to give us zero. So, weβve shown that all of our prerequisites for LβHΓ΄pitalβs rule are true.

By LβHΓ΄pitalβs rule, instead of evaluating the limit given to us in the question, we can instead evaluate the limit as π₯ approaches zero of π prime of π₯ divided by π prime of π₯. Thatβs the limit as π₯ approaches zero of the sin of π₯ divided by π to the power of π₯ minus one. And we know we can evaluate the sin of π₯ and π to the power of π₯ minus one by direct substitution. So, we can attempt to evaluate the limit as π₯ approaches zero of their quotient by direct substitution.

Substituting π₯ is equal to zero, we get the sin of zero divided by π to the zeroth power minus one. And since the sin of zero is equal to zero and π to the zeroth power is equal to one, we see that this evaluates to give us the indeterminate form zero divided by zero. So, it might seem like using LβHΓ΄pitalβs rule has failed us. However, we can see that we try to calculate the limit as π₯ approaches zero of the quotient to two functions and again ended up with the indeterminate form of zero over zero. This could give us the idea of trying to use LβHΓ΄pitalβs rule on this limit instead. In other words, we apply LβHΓ΄pitalβs rule twice.

So, we want to try LβHΓ΄pitalβs rule with π equal to zero, the function in our numerator the sin of π₯, and the function in our denominator π to the power of π₯ minus one. So, we wrote in LβHΓ΄pitalβs rule with our functions π prime of π₯ and π prime of π₯. We get if π prime of π₯ and π prime of π₯ are differentiable functions and π double prime of π₯ is not equal to zero around π₯ is equal to zero. Although π double prime of zero is allowed to be equal to zero. And the limit as π₯ approaches zero of π prime of π₯ and the limit as π₯ approaches zero of π prime of π₯ are equal to zero. Then the limit as π₯ approaches zero of π prime of π₯ divided by π prime of π₯ is equal to the limit as π₯ approaches zero of π double prime of π₯ divided by π double prime of π₯.

Weβll do this in exactly the same way we did before. First, weβll find expressions for π double prime of π₯ and π double prime of π₯. Letβs start with π double prime of π₯. Thatβs the derivative of the sin of π₯, which is just equal to the cos of π₯. Next, we want to find an expression for π double prime of π₯. Thatβs the derivative of π to the power of π₯ minus one, which is just π to the power of π₯. And this shows us that both π prime and π prime are differentiable. Next, we need to show that π double prime of π₯ is not equal to zero around our value of π₯ is equal to zero. And we can see that π prime of π₯ is equal to π to the power of π₯, which we know is greater than zero for all values of π₯, so in particular is not equal to zero.

The last thing we need is the limit as π₯ approaches zero of π prime of π₯ and the limit as π₯ approaches zero of π prime of π₯ are both equal to zero. And again, we already showed this was true. When we attempted to evaluate the limit of π prime of π₯ divided by π prime of π₯ with direct substitution, the numerator of this limit is just the limit as π₯ approaches zero of π prime of π₯. And the denominator of this calculation was the limit as π₯ approaches zero of π prime of π₯. And when we evaluated this, we got the indeterminate form of zero divided by zero. So, weβve already shown that this is true.

So, weβve shown all of the prerequisites for LβHΓ΄pitalβs rule are true. So, letβs apply it again. We have the limit as π₯ approaches zero of π prime of π₯ divided by π prime of π₯ is equal to the limit as π₯ approaches zero of π double prime of π₯ divided by π double prime of π₯. Thatβs the limit as π₯ approaches zero of the cos of π₯ divided by π to the power of π₯. And again, we can attempt to evaluate this by direct substitution. Substituting π₯ is equal to zero, we get the cos of zero divided by π to the zeroth power. And we see the cos of zero is equal to one and π to the zeroth power is also equal to one. So, we can evaluate this limit to just give us one.

So, weβve shown the limit as π₯ approaches zero of one minus the cos of π₯ divided by π to the power of π₯ minus π₯ minus one is just equal to one.