### Video Transcript

Determine the limit as π₯ approaches zero of two π to the power of π₯ minus three times the sin of π₯ minus two all divided by negative two π₯.

Weβre asked to determine the limit as π₯ approaches zero of the quotient of two functions. And we can see the function in our numerator is continuous since itβs the sum of an exponential function, a trigonometric function, and a constant. And the function in our denominator is also continuous because itβs a linear function. This means we can attempt to evaluate this limit by using direct substitution. So weβll substitute π₯ is equal to zero into the function inside of our limit. This gives us two times π to the zeroth power minus three times the sin of zero minus two all divided by negative two multiplied by zero.

And if we evaluate this expression, we will get the indeterminant form of zero divided by zero. Remember, this does not mean our limit is undefined. It means we could not determine our limit by using this method. This means weβre going to need to try and evaluate this using a different method. And because the function inside of our limit is the quotient of two functions, both of which we know how to differentiate, we could try using LβHΓ΄pitalβs rule.

So weβll start by recalling LβHΓ΄pitalβs rule. This tells us if π and π are differentiable and π prime of π₯ is not equal to zero around some constant value of π, although π prime of π is allowed to be equal to zero and both the limit as π₯ approaches π of π of π₯ and the limit as π₯ approaches π of π of π₯ are equal to zero, then the limit as π₯ approaches π of π of π₯ divided by π of π₯ will be equal to the limit as π₯ approaches π of π prime of π₯ divided by π prime of π₯.

The first thing we want to do is determine our functions π and π and our value of π. Letβs start with our value of π. We can see our limit is as π₯ is approaching zero. This means weβll set our value of π equal to zero, and weβll update LβHΓ΄pitalβs rule so that π is equal to zero. Next, weβll set π of π₯ to be the function in the numerator, thatβs two π to the power of π₯ minus three times the sin of π₯ minus two, and π of π₯ to be the function in our denominator. Thatβs negative two π₯.

Now weβre ready to check our prerequisites for LβHΓ΄pitalβs rule. First, we need to check that our functions π and π are differentiable around π₯ is equal to zero. And we can do this directly from the definitions of π of π₯ and π of π₯. Letβs start with π of π₯. We can see itβs the sum or difference of an exponential function, a trigonometric function, and the constant function. And all three of these are differentiable for all real values of π₯. This means π of π₯ will be differentiable for all real values of π₯ and so, in particular, is differentiable around π₯ is equal to zero. Similarly, π of π₯ is a linear function, and we know all linear functions are differentiable for all real values of π₯.

The next thing we need to show is that π prime of π₯ is not equal to zero around π₯ is equal to zero, although π prime of zero is allowed to be equal to zero. And the easiest way to do this is to find an expression for π prime of π₯. Thatβs the derivative of negative two π₯ with respect to π₯, which will just be the coefficient of π₯, which is negative two. And we can of course see this is not equal to zero for any value of π₯. Next, we could show directly the limit as π₯ approaches zero of π of π₯ and the limit as π₯ approaches zero of π of π₯ are equal to zero by using direct substitution.

However, this is not necessary. Consider the numerator and the denominator when we evaluated our original limit by using direct substitution. By looking at just our numerator, we can see we evaluated the limit as π₯ approaches zero of π of π₯ by using direct substitution. We showed this was equal to zero. And in our denominator, we showed the same thing. We showed the limit as π₯ approaches zero of π of π₯ is equal to zero. So both of these limits are equal to zero. And this means we can use LβHΓ΄pitalβs rule.

But to use LβHΓ΄pitalβs rule, we see we need an expression for π prime of π₯. So before we do anything, weβre going to need to differentiate π of π₯ with respect to π₯. And we can just do this term by term. We get π prime of π₯ is equal to two π to the power of π₯ minus three times the cos of π₯. Weβre now ready to apply LβHΓ΄pitalβs rule. We get the limit given to us in the question is equal to the limit as π₯ approaches zero of π prime of π₯ divided by π prime of π₯. Weβll substitute in our expressions for π prime of π₯ and π prime of π₯. This gives us the limit as π₯ approaches zero of two π to the power of π₯ minus three times the cos of π₯ all divided by negative two.

And we can see this is a continuous function, so we can attempt to evaluate this by using direct substitution. Substituting in π₯ is equal to zero, we get two π to the zeroth power minus three times the cos of zero all divided by negative two. And we can simplify this. π to the zeroth power and the cos of zero are both equal to one. So this simplified to give us two minus three divided by negative two, which is negative one over negative two, which is, of course, equal to one-half. Therefore, by using LβHΓ΄pitalβs rule, we were able to determine the limit as π₯ approaches zero of two π to the power of π₯ minus three sin of π₯ minus two all divided by negative two π₯ is equal to one-half.