I want to evaluate the following limit without using the L'Hopital rule : $$ \lim\limits_{x\rightarrow 0^+}\frac{e^{x\ln(x)}-1}{x}$$ I know the answer is $-\infty$. I can demonstrate that graphically and by using the L'Hopital rule. Any hint would be appreciated and thanks.
$\begingroup$
$\endgroup$
2
-
3$\begingroup$ Multiply and divide by $\log x$ $\endgroup$– Paramanand Singh ♦Nov 16, 2018 at 14:40
-
$\begingroup$ e^(xln(x)) = x^x might help $\endgroup$– mbeckishNov 16, 2018 at 20:55
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Note that for $x\log(x)<1$,
$$e^{x\log(x)}\le \frac1{1-x\log(x)}$$
whereby we see that for $x\le 1$
$$\begin{align} \frac{e^{x\log(x)}-1}{x}&\le \frac{\log(x)}{1-x\log(x)}\\\\ &\le \frac{e}{e+1}\,\log(x) \end{align}$$
Inasmuch as $\log(x)\to -\infty$, we find that
$$\lim_{x\to 0^+}\frac{e^{x\log(x)}-1}{x}=-\infty$$
answered Nov 16, 2018 at 16:17
$\begingroup$
$\endgroup$
HINT
The key point is that $x\log x \to 0$ then
$$\frac{e^{x\ln(x)}-1}{x}=\frac{e^{x\ln(x)}-1}{x\ln(x)}\ln x$$
