Using L'Hôpital's rule on $\lim_{x \to 0}(xe^{\frac{1}{x}}-x) $

Question

As stated, we are supposed to use L'Hôpital's rule on:

$$\lim_{x \to 0}(xe^{\frac{1}{x}}-x) $$

We can write this function as:

$$\frac{e^{\frac{1}{x}}-1}{\frac{1}{x}}=\frac{e^{\frac{1}{x}}-1}{x^{-1}}$$

I'm a bit uncertain of the derivative of $e$ to the power of a fraction but I guess the numerator can be written as: $$e^{x*\frac{1}{x^2}}$$

So the derivative would be:

$$\frac{1}{x^2}e^{x*\frac{1}{x^2}}=\frac{1}{x^2}e^\frac{1}{x}$$

The derivative of the denominator should simply be:

$$-1*x^{-2}=-\frac{1}{x^2}$$

Thusly

$$\frac{f(x)}{g(x)}=\frac{f'(x)}{g'(x)}=\frac{\frac{1}{x^2}e^\frac{1}{x}}{-\frac{1}{x^2}}=-\frac{x^2}{x^2}*e^\frac{1}{x}$$

As x tends to 0, $e^{\frac{1}{x}}$ tends to infinity, and $-\frac{x^2}{x^2}$ should tend to...-1 I guess since the denominator will always equal the numerator. In any case, this term should always be negative when $x \neq 0$, so x should equal negative infinity.

That's not the answer however, $x$ should equal positive infinity, how do I reach this result?

Answer 1

Your mistake was the derivative of $e^{\frac{1}{x}}$. Simply by using the chain rule the derivative is $e^{\frac{1}{x}}(-\frac{1}{x^2})$. So it has a negative sign. The rest of your solution is correct, so with the change of the sign the answer really becomes positive infinity.

Oh, and I assume the limit should be $x\to 0^{+}$, because if $x\to 0^{-}$ then $e^{\frac{1}{x}}\to 0$, so then the answer is really different.

Answer 2

Let $\newcommand{\euler}{\mathrm{e}}u=\dfrac{1}{x}$ and apply L'Hospital for $$\lim_{u \to \infty}\dfrac{\euler^u-1}{u}$$

Answer 3

We have that for $x\to 0^+$ by $y=1/x \to \infty$ by l’Hopital rule we have

$$\lim_{x\to 0^+} xe^{1/x}-x=\lim_{y\to +\infty} \frac{e^y-1}{y}=\lim_{y\to +\infty} e^y= \infty$$

but for $x\to 0^-$ by $y=1/x \to -\infty$

$$\lim_{x\to 0^-} xe^{1/x}-x=\lim_{y\to -\infty} \frac{e^y-1}{y}=0$$

therefore the given limit doesn’t exist.