Question

${lim}_{x\to 0^{+}}xlo{g}_{e}x$ will be equal to ___

Answer 1

We can observe that

$Asx\to 0^{+},lo{g}_{e}x\to -\infty$

So if we write the above expression as $\frac{lo{g}_{e}x}{\frac{1}{x}}$ then,

$asx\to 0^{+},\frac{1}{x}\to \infty ;\frac{lo{g}_{e}x}{\frac{1}{x}}reducesto\left(\frac{\infty }{\infty }\right)form.$

So we can use L’Hospital Rule. We are doing all this because direct substitution of x as zero is not giving us the limit.

So L’Hospital Rule states that ${lim}_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}={lim}_{x\to a}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

So for $\frac{lo{g}_{e}x}{\frac{1}{x}}$ taking derivative of numerator and denominator we get $\frac{\frac{1}{x}}{-\frac{1}{x^2}}$

$So{lim}_{x\to 0^{+}}\frac{lo{g}_{e}x}{\frac{1}{x}}={lim}_{x\to 0^{+}}\frac{\frac{1}{x}}{-\frac{1}{x^2}}={lim}_{x\to 0^{+}}(-x)=0$ which is the correct answer.