Limit of $x\ln x$ as $x$ approaches $0^+$

Question

How could I solve the limit in this form $ \frac{x}{x+1}$ using l'Hospital's rule?

I know how to solve it in this way: $\frac{\ln x}{\frac{1}{x}}$

Thanks

Answer 1

You can try using l'Hôpital's theorem: $$ \lim_{x\to0}x\ln x= \lim_{x\to0}\frac{x}{1/\ln x}\overset{\text{(H)}}{=} \lim_{x\to0}\frac{1}{(-1/(\ln x)^2)(1/x)}= \lim_{x\to0}-x(\ln x)^2 $$ This goes nowhere, if you're adamant into transforming the expression into a limit of the form $0/0$: the next step will take you to $$ \lim_{x\to0}\frac{1}{2}x(\ln x)^3 $$ and so on.

It's like being inside a well; you have two directions: down or up. Which one do you choose?

Answer 2

Rewrite it like this

$$\lim_{x\to 0^+} x\ln x = \lim_{x\to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0 $$

EDIT: Your question is vague. Do you want to turn it into $\frac{0}{0}$ instead of $\frac{\infty}{\infty}$? That's not possible, as egreg pointed out.

Answer 3

If $x = e^{-y}$, we want $\lim_{y \to \infty} ye^{-y}$.

$ye^{-y} =\frac{y}{e^y} $.

But, for $y > 0$, $e^y >1+y+y^2/2 >y^2/2 $, so $\frac{y}{e^y} <\frac{y}{y^2/2} = \frac{2}{y} \to 0 $ as $y \to \infty$.

In general, from the power series for $e^y$, for $y > 0$, $e^y > \frac{y^{n+1}}{(n+1)!}$ so $\frac{y^n}{e^y} <\frac{(n+1)!}{y} \to 0$ as $y \to \infty$.