Video Transcript
Let 𝑓 of 𝑥 equal tan two 𝑥 plus cos 𝑥 minus 𝑥 minus one all over 𝑒 to the power of three 𝑥 minus three 𝑥 minus one. Which of the following is true? One, The limit as 𝑥 tends to zero of 𝑓 of 𝑥 exists. Two, 𝑓 of 𝑥 is continuous at 𝑥 equal zero. And three, 𝑓 of 𝑥 is differentiable at 𝑥 equals zero.
Let’s look at these three statements one by one. The first statement says that the limit as 𝑥 tends to zero of 𝑓 of 𝑥 exists. Our function 𝑓 of 𝑥 is a rational function. So let’s call the numerator 𝑔 of 𝑥 and the denominator ℎ of 𝑥. We want to know the behavior of 𝑓 of 𝑥 as 𝑥 tends to zero or as 𝑥 approaches zero, that is, the limit as 𝑥 tends to zero of 𝑓 of 𝑥, which is equal to the limit as 𝑥 tends to zero of 𝑔 of 𝑥 over ℎ of 𝑥. Let’s look first at the numerator. As 𝑥 tends to zero, the tangent of two 𝑥 also tends to zero. The cos of 𝑥 tends to one. 𝑥 itself tends to zero. And we’re left with minus one at the end. So in the numerator, we have zero plus one minus zero minus one, and that’s equal to zero.
If we look next to the denominator 𝑒 to the power of three 𝑥 tends to 𝑒 to the power of zero. And 𝑒 to the power of zero is equal to one. Negative three 𝑥 tends to zero as 𝑥 tends to zero. And again, we have negative one at the end. The limit as 𝑥 tends to zero of the denominator is then one minus zero minus one, and this is equal to zero. So we have that the limit as 𝑥 tends to zero of 𝑓 of 𝑥 which is equal to 𝑔 of 𝑥 over ℎ of 𝑥 is equal to zero over zero. This means that the limit of the function 𝑓 of 𝑥 as 𝑥 tends to zero is undefined. And a limit of this type is called an indeterminate form of type zero over zero.
Because our limit is of this type, we may be able to use L’Hôpital’s rule. This says that if 𝑔 and ℎ are differential and ℎ prime of 𝑥 does not equal zero near 𝑎, except possibly at 𝑎, and we have an indeterminate form of type zero over zero. Then the limit as 𝑥 tends to 𝑎 of 𝑔 of 𝑥 over ℎ of 𝑥 is equal to the limit as 𝑥 tends to 𝑎 of 𝑔 prime of 𝑥 over ℎ prime of 𝑥. This is true if the right-hand side limit exists or is plus or minus infinity. The most important thing we need to check before applying L’Hôpital’s rule is the condition that ℎ prime of 𝑥 does not equal zero near 𝑎, except possibly at 𝑎. In our case, ℎ of 𝑥 is 𝑒 to the three 𝑥 minus three 𝑥 minus one. So ℎ prime of 𝑥, the derivative of ℎ with respect to 𝑥, is equal to three 𝑒 to the three 𝑥 minus three by the chain rule. And this is equal to three times 𝑒 to the power of three 𝑥 minus one.
Now, we’re looking at the limit as 𝑥 tends to zero. So in L’Hôpital’s rule 𝑎 is equal to zero. If ℎ prime of 𝑥 does not equal to zero around 𝑥 equals zero, then we can apply L’Hôpital’s rule. To look at the behavior around 𝑥 equals zero, we can draw a graph of ℎ prime of 𝑥. We can see that the function actually does go through the origin. But if we zoom in at the origin, we can see that as 𝑥 approaches zero from either side, ℎ prime of 𝑥 is not equal to zero so that our function ℎ prime of 𝑥 satisfies the condition for L’Hôpital’s rule. We have ℎ prime of 𝑥 is equal to three times 𝑒 to the three 𝑥 minus one. So now if we find 𝑔 prime of 𝑥, we can use L’Hôpital’s rule. 𝑔 of 𝑥 is equal to tan two 𝑥 plus cos 𝑥 minus 𝑥 minus one. And this is the numerator in our function 𝑓 of 𝑥.
Differentiating 𝑔 with respect to 𝑥, we then have the derivative of tan two 𝑥 is equal to two sec squared two 𝑥. The derivative of cos 𝑥 is minus sin 𝑥 and the derivative of minus 𝑥 is minus one. Remember, we’re looking for the limit as 𝑥 tends to zero of 𝑔 prime of 𝑥 over ℎ prime of 𝑥. And this is equal to the limit as 𝑥 tends to zero of two sec squared two 𝑥 minus sin 𝑥 minus one over three times 𝑒 to the three 𝑥 minus one. Remember that sec 𝑥 is one over cos 𝑥. So two sec squared two 𝑥 is two over cos squared two 𝑥. And as 𝑥 tends to zero, cos 𝑥 tends to one. So two over cos squared two 𝑥 tends to two over one which is equal to two.
As 𝑥 tends to zero sin 𝑥 tends to zero. And we know that in the denominator, 𝑒 to the three 𝑥 tends to one. So the limit as 𝑥 tends to zero of 𝑔 prime of 𝑥 over ℎ prime of 𝑥 is equal to two minus zero minus one over one minus one, which is equal to zero. And our limit is one divided by zero. Because the denominator tends to zero as 𝑥 tends to zero but the numerator does not, we can say that the limit of 𝑓 of 𝑥 as 𝑥 tends to zero is undefined and, therefore, does not exist. Therefore, statement one that the limit as 𝑥 tends to zero of 𝑓 of 𝑥 exists is not true.
Let’s look now to statement number two. This says that 𝑓 of 𝑥 is continuous at 𝑥 equal zero. Now, we know that at 𝑥 equals zero 𝑓 of 𝑥 is equal to tan two zero plus cos of zero minus zero minus one all over 𝑒 to the power of three times zero minus three times zero minus one. We know that the tangent of zero is zero. We know that the cos of zero is equal to one. 𝑒 to the power of zero is one. And three times zero is zero. So 𝑓 at 𝑥 equals zero is equal to zero plus one minus zero minus one over one minus zero minus one. And we know that this is equal to zero over zero. Hence, 𝑓 is undefined and, therefore, not continuous at 𝑥 equal to zero. So statement number two is not true.
Now let’s look at statement number three. This says that 𝑓 of 𝑥 is differentiable at 𝑥 equal to zero. But if 𝑓 is not continuous at 𝑥 equal to zero, then it cannot be differentiable at 𝑥 equal to zero. So statement three is also false. Therefore, all three statements one, two, and three are false.
