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Sorry I haven't figured out how to properly write up questions on this yet. I have to get the limit as $x$ goes to $1$, of the function $- (x-1)/\log(x)$ without using L’Hospital’s rule, any suggestions?

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Setting $x-1=t$, you have to find the limit of $-\dfrac{t}{\ln(1+t)}$ as $t$ tends to $0$, and a well-known high school result asserts that $$\lim_{t\to 0}\frac{\ln(1+t)}t=1,$$ i.e. the slope of the curve $y=\ln x$ at the point $(1,0)$ is equal to $1$.

Bernard
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  • "Well-known highschool result" Also known as "the definition of the derivative of $\ln(x)$ at $x=1$". – Arthur Dec 13 '19 at 21:29
  • Yes, there are several variants in formulation. Even more directly: $;\dfrac{\ln x}{x-1}\to 1$ as $x\to 1$. Every well-bred young people should know these variants. – Bernard Dec 13 '19 at 21:33
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You use the definition of the derivative:

$$ \left.\frac{d}{dt}\log(t)\right|_{t=1} = \lim_{x \to 1} \frac{\log(x) - \log(1)}{x-1}. \tag{$*$} $$

From there, how you compute the derivative depends on how you define $\log$. Two common definitions are:

  1. $\log(x) = \displaystyle \int_{1}^x \frac1t \;dt$. Then you compute $(*)$ using the Fundamental Theorem of Calculus.
  2. $\log(x)$ is the inverse function to $e^x$. Then you compute $(*)$ using the formula $$ \frac{d}{dt}f^{-1}(y) = \frac{1}{f'(x)}\, \text{ where } y = f(x). $$ In this case, the derivative is $1/e^0 = 1$.
J.G.
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Trevor Gunn
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    I've changed your notation to emphasise we differentiate the function before setting $t=1$. To my eye, $\frac{d}{dt}\log 1=0$. – J.G. Dec 13 '19 at 23:01
  • Would it be possible for me to use taylors theorem? I tried using it to get an approximation of log(x) and ended up with the right answer but not sure if that’s a suitable method. –  Dec 14 '19 at 23:15
  • @oliviamclo Taylor's theorem is just l'Hôpital's rule with extra steps. – Trevor Gunn Dec 15 '19 at 03:07
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Substitute $x=e^y$, so the limit becomes $$ \lim_{y\to0}-\frac{e^y-1}{y}=-1 $$ If you aren't allowed to use l’Hôpital, you surely are allowed to use some “well-known” limits, aren't you?

egreg
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