Video: APCALC04AB-P1A-Q23-364102370513

Find the lim(β„Ž β†’ 0) (ln(π‘₯ + 5 + β„Ž) βˆ’ ln(π‘₯ + 5))/β„Ž.

03:23

Video Transcript

Find the limit as β„Ž approaches zero of the natural log of π‘₯ plus five plus β„Ž minus the natural log of π‘₯ plus five all over β„Ž.

When working with limits, it’s always sensible to begin by checking whether we can find the limit by using direct substitution. In fact, in this case, if we were to use direct substitution, we would have some number divided by zero. And we know that’s undefined. So, instead, we’re going to need to find some way to manipulate our function. In fact, it’s really useful to recall this specific law of logs, the natural log of π‘Ž minus the natural log of 𝑏 is equal to the natural log of π‘Ž divided by 𝑏. And this means we can change the numerator of our fraction and write it as the natural log of π‘₯ plus five plus β„Ž over π‘₯ plus five.

We’ll rewrite this as one over β„Ž times the natural log of π‘₯ plus five plus β„Ž over π‘₯ plus five. And then, we’re going to simplify this fraction. We’re going to write it as π‘₯ plus five over π‘₯ plus five plus β„Ž over π‘₯ plus five. And this is great because we know that π‘₯ plus five over π‘₯ plus five is simply one. Okay, so we have the limit as β„Ž approaches zero of one over β„Ž times the natural log of one plus β„Ž over π‘₯ plus five. We’re going to perform a rather strange step. We’re going to let 𝑒 be equal to β„Ž over π‘₯ plus five. And we see that as β„Ž approaches zero, 𝑒 itself approaches zero.

We rearrange this to make β„Ž the subject so that β„Ž is equal to 𝑒 times π‘₯ plus five. And then, we change our limit. So, we have the limit as 𝑒 approaches zero of one over 𝑒 times π‘₯ plus five times the natural log of one plus 𝑒. π‘₯ plus five or one over π‘₯ plus five is completely independent of 𝑒. So, we can take one over π‘₯ plus five outside of our limit. And then, we recall a second law of logs. And that is π‘Ž times the natural log of 𝑏 can be written as the natural log of 𝑏 to the power of π‘Ž. And so, we rewrite our limit. It’s the natural log of one plus 𝑒 to the power of one over 𝑒.

And then, we have quite a strange step. The limit of the natural log of a function is equal to the natural log of the limit of that function. So, we take the limit inside the natural log function. And we see that we’re looking for one over π‘₯ plus five times the natural log of the limit as 𝑒 approaches zero of one plus 𝑒 to the power of one over 𝑒. And we complete one more substitution. We let 𝑣 be equal to one over 𝑒 so that when 𝑒 approaches zero, 𝑣 approaches ∞.

By rewriting 𝑣 equals one over 𝑒 as 𝑒 equals one over 𝑣, our limit becomes the limit as 𝑣 approaches ∞ of one plus one over 𝑣 to the power of 𝑣. But this part here is simply the definition of 𝑒. So, let’s clear some space and complete the final step. Our limit becomes one over π‘₯ plus five times the natural log of 𝑒. Now, the natural log of 𝑒 is simply one. So, our limit is equal to one over π‘₯ plus five.

Now, in fact, this isn’t the only method we could’ve used. We could, alternatively, have recalled the definition of the derivative. 𝑓 prime of π‘₯ is equal to the limit as β„Ž approaches zero of 𝑓 of π‘₯ plus β„Ž minus f of π‘₯ all over β„Ž. We then see if we compare this with our original question, we can let 𝑓 of π‘₯ be equal to the natural log of π‘₯ plus five. And so, the limit in our question is simply asking us to evaluate the derivative of 𝑓 of π‘₯, the derivative of the natural log of π‘₯ plus five with respect to π‘₯, which of course we know to be equal to one over π‘₯ plus five.

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