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i am trying to draw the graph of $$f(x)=\frac{e^x-1}{x}$$

its domain is $\mathbb{R}$-$0$ but its limit is $1$ in the neighbourhood of zero.

Also $$lim_{x\to -\infty}f(x)=0$$ so negative X axis is asymptote.

also $$\frac{df(x)}{dx}=\frac{(x-1)e^x+1}{x^2}$$

so $f(x)$ is increasing in $(0 \: \infty)$ and decreasing in $(-\infty \:0)$

now the ambiguity is $f(-\infty) \to 0$ and $f(0^-) \to 1$ but $f(x)$ is decreasing in negative interval.

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  • $\begingroup$ The derivative is positive for $x<0\lor x>0$ so $f$ is increasing in both intervals $\endgroup$ – Alessandro Codenotti Sep 12 '15 at 16:04
  • $\begingroup$ Concerning the graph of $f(−x)$ (the horizontal 'mirror' of your function) as well as of its inverse (the 'bisector mirror') see this thread. $\endgroup$ – Raymond Manzoni Sep 12 '15 at 16:30
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There is no ambiguity. The mistake is when you claim that $f$ is decreasing in$(-\infty,0)$ since for $x<0$ you still have $f'(x)=\frac{(x-1)e^x+1}{x^2}>0$.

This is because of the well-known $e^x \ge x+1$ for all $x$, in particular $1-x \le e^{-x}$ i.e. $(1-x)e^x \le 1$ and hence $(x-1)e^x+1 \ge 0$.

So $f$ is increasing everywhere which goes along with $f(-\infty)<f(0)$

See also here for the graph: www.wolframalpha.com/input/?i=%28e^x-1%29%2Fx

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