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

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}

{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}