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hibz 660 · Nov 9, 2008, 10:28 AM

Evaluate the following limit:-

xlnx
where (x approuch to 0)
I tried to answer this question by l'hopital rule and got 1 as the answer... is it correct? please need to know if I solved it correctly or not :confused:and how to solve it by l'hopital rule.. thx:D

galactus · Nov 9, 2008, 01:40 PM

To be formal, we can use the Squeeze theorem.

Recall the definition $\int_{1}^{x}\frac{dy}{y}=ln(x)$

For $y\geq 1$ we get $y\geq \sqrt{y}$

$\Rightarrow \frac{1}{y}\leq \frac{1}{\sqrt{y}}$

$\Rightarrow \int_{1}^{x}\frac{1}{y}dy\leq$

$\int_{1}^{x}\frac{1}{\sqrt{y}}dy$ .

As $\int_{1}^{t}\frac{1}{\sqrt{y}}dy=2\sqrt{y}|_{1}^{x }=2\sqrt{x}-2$

We get $ln(x)=\int_{1}^{x}\frac{1}{y}dy\leq 2\sqrt{x}-2$

Now, the Squeeze theorem:

For $x\geq 1$ , $0\leq \frac{ln(x)}{x}\leq \frac{2\sqrt{x}-2}{x}$

$0\leq \frac{ln(x)}{x}\leq 2\left(\frac{1}{\sqrt{x}}-\frac{1}{x}\right)$

As $2\left(\frac{1}{\sqrt{x}}-\frac{1}{x}\right)\rightarrow_{x\to \infty}0$

we get $\frac{ln(x)}{x}\rightarrow_{x\to \infty} 0$

Now, let $w=\frac{1}{x}$ :

$\lim_{x\to \infty}\frac{ln(x)}{x}-\lim_{w\to 0^{+}}\frac{ln(\frac{1}<br /> <br /> {w})}{\frac{1}{w}}=\lim_{w\to 0^{+}} -wln(w)$

Since $\frac{ln(x)}{x}\rightarrow_{x\to {\infty}}0$ we get

$\fbox{\lim_{w\to 0^{+}}wln(w)=0}$