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18-01F07-L07

# 18-01F07-L07 - MIT OpenCourseWare http/ocw.mit.edu 18.01...

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MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus, Fall 2007 Transcript – Lecture 7 The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or to view additional material from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Right now, we're finishing up with the first unit, and I'd like to continue in this lecture, lecture seven, with some final remarks about exponents. So what I'd like to do is just review something that I did quickly last time, and make a few philosophical remarks about it. I think that the steps involved were maybe a little tricky, and so I'd like to go through it one more time. Remember, that we were talking about this number ak, which is (1 1/k)^k. And what we showed was that the limit as k goes to infinity of ak was e. So the first thing that I'd like to do is just explain the proof a little bit more clearly than I did last time with fewer symbols, or at least with this abbreviation of the symbol here, to show you what we actually did. So I'll just remind you of what we did last time, and the first observation was to check, rather than the limit of this function, but to take the ln first. And this is typically what's done when you have an exponential, when you have an exponent. And what we found was that the limit here was 1 as k goes to infinity. So last time, this is what we did. And I just wanted to be careful and show you exactly what the next step is. If you exponentiate this fact; you take e to this power, that's going to tend to e ^ 1, which is just e. And then, we just observe that this is the same as ak. So the basic ingredient here is that e ^ ln a = a. That's because the ln function is the inverse of the exponential function. Yes, question? STUDENT: [INAUDIBLE] PROFESSOR: So the question was, wouldn't the log of this be because ak is tending to 1. But ak isn't tending to 1. Who said it was? If you take the logarithm, which is what we did last time, logarithm of ak is indeed k(ln(1 1/k)). That does not tend to 0. This part of it tends to 0, and this part tends to infinity. And they balance each other, times infinity. We don't really know yet from this expression, in fact we did some cleverness with limits and derivatives, to figure out this limit. It was a very subtle thing. It turned out to be 1. All right?
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