I am trying to determine the limit $\displaystyle{ \lim_{x \to 0^-}{\frac{-e^{1/x}}{x}}}$. Plugging in $x$ directly, yields $0/0$ which is indeterminate. Applying L'Hopitals rule does not simplify the fraction, in fact the result is even more complicated and looks like a dead end, i.e. $\displaystyle{ \lim_{x \to 0^-}{\frac{-e^{1/x}}{x}} = \lim_{x \to 0^-}{\frac{e^{1/x}}{x^2}}}$. So I tried taking the natural log, i.e. $\displaystyle{ y = \lim_{x \to 0^-}{\frac{-e^{1/x}}{x}} }$

$\displaystyle{ \ln(y) = \lim_{x \to 0^-}{ \ln \left( \frac{e^{1/x}}{-x} \right) } = \lim_{x \to 0^-}{ [1/x - \ln(-x) ]} = \lim_{x \to 0^-}{ \frac{1-x\ln(-x)}{x} } }$

Applying L'Hopitals Rule here gives me $\displaystyle{ \lim_{x \to 0^-}{ -1-\ln(-x) } = +\infty }$ which is wrong; it should be $-\infty$.

I think I am making a very simple mistake but I don't see it.

Can anyone help?

  • 1
    $\begingroup$ Try the variable change $u = \frac{1}{x},$ noting that $\displaystyle{ \lim_{x \to 0^-}}$ gets converted to $\displaystyle{ \lim_{u \to -\infty}}.$ $\endgroup$ – Dave L. Renfro Aug 4 '14 at 19:58
  • $\begingroup$ For $x < 0$, we find that $-e^{1/x} < 0$, hence the ratio of these two quantities is positive. How then would the limit be $-\infty$? $\endgroup$ – David K Aug 4 '14 at 19:59
  • $\begingroup$ @DavidK: The limit of the log is to $-\infty$, not of the quantity itself. $\endgroup$ – Semiclassical Aug 4 '14 at 20:04
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    $\begingroup$ After you do what I suggested, rewrite the quotient so that the exponential is in the denominator with $u$ in the numerator. $\endgroup$ – Dave L. Renfro Aug 4 '14 at 20:05
  • $\begingroup$ @Semiclassical: never mind, I misread the line where $-\infty$ appeared. $\endgroup$ – David K Aug 4 '14 at 20:07

The expression $\frac{1-x\ln(-x)}{x}$ is not one of the indeterminate forms to which L'Hospital's Rule can be applied. It is not really an indeterminate form: the top approaches $1$, and the bottom approaches $0$. That we need to work a bit to show that the top approaches $1$ does not change things.

But once we are there life is simple. The bottom approaches $0$ through negative values, so the expression becomes very large negative as $x\to 0^-$.

It follows that the original expression has limit $0$.

  • $\begingroup$ I see that now. Thank you! $\endgroup$ – 017 Aug 4 '14 at 22:33
  • $\begingroup$ You are welcome. Easy slip to make, once one has gotten into a differentiation mood. $\endgroup$ – André Nicolas Aug 4 '14 at 22:38

$ \lim_{x \to 0^{-}}\bigg(\frac{1}{(-x)} \bigg)e^{\frac{1}{x}} = \lim_{t \to \infty} t e ^{-t} = \lim_{t \to \infty} e^{\log t e^{-t}} = \lim_{t \to \infty} e^{\log t - t } =0$


$$\frac{e^{\frac{1}{x}}}{x}=\frac{1}{x}+o\left(\frac{1}{x}\right) \text{ if }\ x\to \infty$$

For more detail, $e^y=1+o(1)$ if $y\to 0$, and so $ye^y=y+o(y)$ if $y\to 0$, then replace $y$ by $\frac{1}{x}$ and you'll have the result.


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