anonymous
  • anonymous
In Unit 1 exercises, 1I-3, it says use d/dx ln(1+x) at x=0 equals 1. Is that the same thing as d/dx ln(x) at x=1 equals 1? And isn't this what is actually needed in this exercise? So, is the idea that we're supposed to figure out that given the first, we've got the second? Seems to me that this could be clearer, but maybe I just don't got the MIT stuff....
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chestercat
  • chestercat
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phi
  • phi
yes, it is the same thing as d/dx ln(x) at x=1 maybe they put it the first way to jog your memory about the definition of the derivative as a limit: \[ \frac{d f(x)}{dx} = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\] using that definition: \[ \frac{d \ln(x)}{dx} \Bigg|_{x=1}= \lim_{h \rightarrow 0} \frac{\ln(1+h)-\ln(1)}{h}=1\] the limit of a sum is the sum of the limits, so \[ \lim_{h \rightarrow 0} \frac{\ln(1+h)}{h}- \lim_{h \rightarrow 0} \frac{\ln(1)}{h}=1 \] ln(1) is 0, and h never equals zero, so the second limit is 0 and we get \[ \lim_{h \rightarrow 0} \frac{\ln(1+h)}{h}=1 \] or \[ \lim_{h \rightarrow 0} \frac{1}{h} \ln(1+h) =1 \]
phi
  • phi
let h= 1/n , so the limit in terms of n is \[ \lim_{n\rightarrow \infty} n \ \ln\left(1+\frac{1}{n} \right) = 1\] bring the n inside the log: \[ \lim_{n\rightarrow \infty} \ \ln\left(1+\frac{1}{n} \right)^n = 1\] make each side the exponent of the base e \[ e^{\lim_{n\rightarrow \infty} \ \ln\left(1+\frac{1}{n} \right)^n}= e \\ \lim_{n\rightarrow \infty} e^{ \ln\left(1+\frac{1}{n} \right)^n}= e\\ \lim_{n\rightarrow \infty}\left(1+\frac{1}{n} \right)^n=e \]
anonymous
  • anonymous
Thank you for your replies.... It was actually mostly clear, but for the odd thing about using that particular d/dx. Another thing. This exercise needed the rule that 'lim f(g(x)) = f(lim(g(x))' or in this case 'lim [e^f(x)] = e^lim [f(x)]', which is not proved anywhere, not that it can't be Googled, but it is interesting that they took the time to prove the product and quotient rules, but not this rule that is, I think, less intuitive.

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