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I have been formulating clues about how a graph of $\frac{\ln x}{x-1}$ might look like. Here is what I found:

1) There are no stationary points

2) The graph at a first glance does not appear to be continuous, as $x \neq 1$

3) The domain is $x \in R^+$

4) The function is above the $x$-axis

Now, the struggle comes when I consider the limits. And the lack of this final clue, means I cannot sketch the graph:

$$\lim_{x \rightarrow 1^+} f(x) = +\infty ?$$ $$\lim_{x \rightarrow 1^-} f(x) = + \infty \ or \ 0?$$ $$\lim_{x \rightarrow +\infty} f(x) = 0$$ $$\lim_{x \rightarrow 0^+} f(x) = + \infty$$

I know for sure that the first two are wrong, but why, and what should the values be then? It also appears that the curve is continuos actually.. Can't understand how.

The clue that there are no stationary points and the fact that I have double asymptote around 1 and then at 0 are conflicting. Since, my derivatives are correct, the limits must be wrong.

Naz
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  • use L'Hôpital's rule to find $\lim_{x \rightarrow 1} f(x)$ – inspd Sep 21 '15 at 13:18
  • I suppose I have to read that up, I hope it is not too advanced for me – Naz Sep 21 '15 at 13:23
  • someone beat me to it; also x=0 and y=0 are asymptotes, and the curve will look like a $y=\frac{1}{x}$ graph – inspd Sep 21 '15 at 13:23
  • Wolfram is giving the function under the x-axis. For $2$, its value is something about $-0.6$. But I can't understand why because $\ln(2)\approx0.692$ – Aditya Agarwal Sep 21 '15 at 13:40
  • Look at the real part in wolfram, not the complex one. – Naz Sep 21 '15 at 14:16
  • @inspd I can't understand why it looks like that, because $x \neq 1$, which implies there should be discontinuity at that point? – Naz Sep 21 '15 at 14:17

1 Answers1

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By the L'Hôpital's rule, $$\lim_{x\to1}\frac{\ln x}{x-1}=\lim_{x\to1}\frac{(\ln x)^\prime}{(x-1)^\prime}=1$$

joefu
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  • Ohh this is beautiful!! So elegant. Need to learn why it works and how was it derived – Naz Sep 21 '15 at 14:49
  • If you don't know L'Hôpital's rule, you can also find the limit by this$$\lim_{x\to1}\frac{\ln x}{x-1}=\lim_{x\to0}\frac{\ln(1+x)}x=\lim_{x\to0}\ln[(1+x)^{1/x}]=\ln[\lim_{x\to0}(1+x)^{1/x}]=\ln e=1$$ – joefu Sep 21 '15 at 16:12