Exponential and Logarithmic Limits

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Suppose a bank pays one hundred percent interest a year, and you put in one rupee. Interest added once a year: you get two rupees. Added twice a year, half each time: one point five, squared, that is two point two five. Ten times a year? About two point five nine. So if the bank adds interest every second, does your one rupee grow without end? Hmm. Keep that question in mind. Today we meet the special number e, and the standard limits of e to the power x and of log. They come up often in JEE, so let's begin.

0:39Where the number e comes from

So, our bank puzzle as a limit. lim, short for limit, with x tends to infinity written below it: the arrow means tends to, and the sideways eight is infinity. Then one plus one over x, in brackets, to the power x: interest added x times a year. Now, the base tends to one, and the power tends to infinity. One to any power is one. So the limit is one, right? No, no, careful! Look at the numbers. x equals two: one point five squared, two point two five. x equals one thousand: one point zero zero one, to the power one thousand, about two point seven one seven; the wavy equals sign means approximately. The base gets closer to one, but the power grows at the same time. Clear? So here is the answer to our puzzle. The limit as x tends to infinity, of one plus one over x, to the power x, is e, about two point seven one eight. So even with interest every second, your one rupee grows only to about e rupees. Not infinite money! Got it? The course also writes it another way. The limit as x tends to zero, of one plus x, to the power one over x, is also e. Tiny x, so the power one over x is huge in size, from either side; still the same e. With me?

2:05Exponential limits: e^x and a^x as x tends to 0

Now, e to the power x, minus one, over x, as x tends to zero; the small x up top is the power. Put x equals zero. What do you get? Zero over zero, exactly! e to the power zero is one, so the top is zero, and the bottom is zero too. Zero over zero form is not an answer; it only says more work is needed. Okay? And the course gives the result. The limit as x tends to zero, of e to the power x minus one, over x, equals one. Near zero, e to the power x minus one behaves just like x. Right? Now replace e by any base, small a. The limit as x tends to zero, of a to the power x minus one, over x, equals l n of a; more on l n in a moment. The bracket says a is greater than zero: the base must be positive. Clear? Here l n a means the natural log of a, log to the base e; the small e below log is the base. We read l n a simply as log a. Quick check: put a equals e in the second result. What is log e? One, yes! So the second result gives back the first. The two formulas agree. Very good!

3:29The log limit, and replacing x by u

Next, the log limit. The limit as x tends to zero, of log of one plus x, over x, equals one. Again zero over zero, because log one is zero, and again the answer is one. Okay? Now the most useful rule of this lesson. In each of these results, x may be replaced by any expression u that tends to zero. e to the power u minus one, over u, tends to one. Log of one plus u, over u, tends to one. But see the condition: the same u on top and in the bottom, and u must tend to zero. Got it? Let us use it on a course example. The limit as x tends to zero, of log of one plus three x, over sine two x. Here u is three x. So what must sit below it? Three x, exactly! So write log of one plus three x, over three x; then two x over sine two x, from our trigonometric limits; and fix the numbers with three over two. The small dot means times. The first two pieces tend to one, so the answer is three over two. Perfect!

4:37Example 1: limits of (e^(3x) - 1)/x and (2^x - 1)/(3^x - 1)

Now, Example one, an easy one from the course, in two parts. First, the limit as x tends to zero, of e to the power three x, minus one, over x. Second, two to the power x minus one, over three to the power x minus one. Pause the video, and try it first. Done? Okay, let's check. The first one looks just like our standard limit. So the answer is one, right? Oh no, careful! Here u is three x, but the bottom is only x. So multiply and divide by three: three times, e to the power three x minus one, over three x. That tends to three times one. The answer is three. Clear? Now the second part, again zero over zero. Divide the top and the bottom by x; the slash means divided by. Now the top tends to log two, and the bottom to log three. So the limit is log two over log three. Okay? And that is log of two to the base three; the small three below log is the base. Change of base rule: both forms are correct. Very good!

5:50Example 2: (12^x - 3^x - 4^x + 1)/x squared as x tends to 0

Now a harder one from the course. The limit as x tends to zero, of twelve to the power x, minus three to the power x, minus four to the power x, plus one, all over x squared. Pause the video and try. Hint: twelve is three times four. Done? Okay, let's check. Put x equals zero: one minus one minus one plus one, zero on top, and zero below. Zero over zero form. With me? Now use the hint. Twelve is three times four, so how can we write twelve to the power x? Three to the power x, times four to the power x, exactly! The power goes to each factor. Now the top has four terms. Can you factorise it? Yes! Take three to the power x common from the first two terms, and minus one from the last two. Then four to the power x minus one is common to both. The top becomes three to the power x minus one, times four to the power x minus one. Okay? Now split x squared as x times x, one x for each bracket. Three to the power x minus one, over x, times four to the power x minus one, over x. Each piece is a standard limit. Right? So the limit is log three, times log four. See the trick? x squared below told us to look for two factors on top. Remember that pattern!

7:22Quick revision: e^u, a^u, ln(1 + u) limits and the number e

So, quick revision. One: when u tends to zero, e to the power u minus one, over u, tends to one; a to the power u minus one, over u, tends to log a; log of one plus u, over u, tends to one. Two: one plus one over x, to the power x, tends to e as x tends to infinity, not to one. Three: same u on top and below, and factorise into standard pieces. Now try the practice problems on this page. All the best!