Video: MATH-DIFF-INT-2018-S1-Q09

Find lim_(π‘₯β†’0) (2^π‘₯ βˆ’ 1)/3π‘₯.

05:13

Video Transcript

Find the limit, as π‘₯ goes to zero, of two to the π‘₯ minus one over three π‘₯.

To start off with, let’s try substituting π‘₯ equals zero into two to the π‘₯ minus one over three π‘₯. We end up with two to the power of zero minus one over three times zero. Two to the power of zero is one. And so, we end up with one minus one over zero, or zero over zero. Now this is undefined. And so, we cannot find our limit this way. So we need to try another method to evaluate our limit. We’ll be using a very useful rule for evaluating limits. And this rule is called L’Hopital’s rule.

In order to use L’Hopital’s rule, there’s a few things that our limit must satisfy. Our limit must be of this form. So that’s the limit as π‘₯ goes to some constant 𝑐 of 𝑓 of π‘₯ over 𝑔 of π‘₯, where 𝑓 and 𝑔 are functions of π‘₯. If we look at our limit, we can see that the constant 𝑐 will be zero, 𝑓 of π‘₯ will be two to the π‘₯ minus one, and 𝑔 of π‘₯ will be the denominator of our fraction, which is three π‘₯. Now, in order to use L’Hopital’s rule, we must have that the limit as π‘₯ goes to 𝑐 of 𝑓 of π‘₯ is equal to the limit as π‘₯ goes to 𝑐 of 𝑔 of π‘₯ is equal to zero or plus or minus infinity.

Now, in our case, we have that the limit of 𝑓 of π‘₯ as π‘₯ goes to zero is the limit as π‘₯ goes to zero of two to the π‘₯ minus one, which is also equal to two to the zero minus one. And since two to the zero is just one, this is equal to zero. And then, the limit of 𝑔 of π‘₯ as π‘₯ goes to zero is the limit as π‘₯ goes to zero of three π‘₯, which is also equal to three times zero, which is also equal to zero. And so, we can see that the limit of 𝑓 of π‘₯ as π‘₯ goes to zero is equal to the limit of 𝑔 of π‘₯ as π‘₯ goes to zero. And they’re both equal to zero. So we can say that our limit is in the correct form for L’Hopital’s rule. And the limit as π‘₯ goes to zero of 𝑓 of π‘₯ is equal to the limit as π‘₯ goes to zero of 𝑔 of π‘₯ is both equal to zero.

The final rule which must be satisfied in order to use L’Hopital’s rule is that 𝑔 prime of π‘₯ must be nonzero for all π‘₯ not equal to 𝑐. 𝑔 prime of π‘₯ is simply the derivative of 𝑔 with respect to π‘₯. Let’s find the derivative of our 𝑔 of π‘₯ with respect to π‘₯. We have that 𝑔 prime of π‘₯ is equal to d by dπ‘₯ of 𝑔 of π‘₯. So, in our case, that’s d by dπ‘₯ of three π‘₯, which is also equal to three. And we can see that three is nonzero for all π‘₯, since three is just a constant. And so, therefore, this last requirement of L’Hopital’s rule is also satisfied.

L’Hopital’s rule tells us that the limit as π‘₯ goes to 𝑐 of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to the Limit as π‘₯ goes to 𝑐 of 𝑓 prime of π‘₯ over 𝑔 prime of π‘₯. Now, in our case, we have that our 𝑓 of π‘₯ is two to the π‘₯ minus one, our 𝑔 of π‘₯ is three π‘₯, our 𝑔 prime of π‘₯ is three. And, now, all we need to find is 𝑓 prime of π‘₯. And 𝑓 prime of π‘₯ is equal to d by dπ‘₯ of 𝑓 of π‘₯. And since our 𝑓 of π‘₯ is two to the π‘₯ minus one, we can substitute this in here. So we need to differentiate two to the π‘₯ minus one with respect to π‘₯.

In order to differentiate two to the π‘₯, we will be using the following rule. We have that d by dπ‘₯ of π‘Ž of π‘₯ is equal to π‘Ž to the π‘₯ ln of π‘Ž. Now, the derivation of this is not necessary to answer this question. However, for those of you who are interested, I’ll leave it on screen for a few seconds so you can pause the video and read through it.

Using this rule for differentiating π‘Ž to the power of π‘₯, we obtain that 𝑓 prime of π‘₯ is equal to two to the π‘₯ times by ln of two. Then, since the minus one term is a constant, this will differentiate to zero. And so, we can leave it out here. We now know 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. So we can use L’Hopital’s rule on the original limit. And we obtain that the limit as π‘₯ goes to zero of two to the π‘₯ minus one over three π‘₯ is equal to the limit as π‘₯ goes to zero of two to the π‘₯ times by ln of two all over three. And then, this is equal to two to the power of zero times by ln of two all over three. And since two to the power of zero is simply one, we obtain that the original limit is equal to ln, or the natural logarithm, of two over three.

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