indeterminate limit where applying L'Hopitals Rules directly doesn't help and using ln gives wrong answer

Question

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?

Answer 1

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 $0$.

Answer 2

$ \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$

Answer 3

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

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.