Calculus problem #2161

ln =1 when limit approches to +0 0(1)=0

Rewrite $x\ln{x}$ as $\frac{\ln{x}}{\frac{1}{x}}$ , then we can apply L'Hôpital's rule to get $-x$ , hence the answer is $0$ .

ln(x)= (x-1) - ((x-1)^2)/2 + ((x-1)^3)/3 -................... each term have only one free term when this free term is multiplied by x in xln(x) then the less term will be x when getting lim when x tends to zero , it will be zero

We have an indetermination $\infty*0$ . The limit can be rewritten as:

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

and we have a $\infty * \infty$ indetermination. Using L'Hopital, we have:

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