$$\lim_{x\to 0^{+}} \frac{1+x\ln x}{x}$$
I think it is an indeterminate form so I applied the L'H rule and did the derivative of numerator and denominator and got limit as x tends to 0 from right [ $\ln (x) + 1$ ] which clearly shows that it depends on $\ln$ function. we know that $\ln$ function is negative infinity on zero plus but when I put this function in WolframAlpha I get positive infinity as the limit of the function. can somebody explain where am I going wrong?
Best Answer
To apply L'Hôpital, you need your limit to be of the form $0/0$ or $\infty/\infty$. Your limit is neither: the numerator converges to $1$ and the denominator converges to $0$. As the limit is one-sided, so $x>0$, the limit equal $+\infty$.
If you don't know the limit of $x\ln x$, one way to do this limit is to substitute $x=1/t$. Then the limit becomes $$ \lim_{t\to\infty}\frac{\ln 1/t}{t}=-\lim_{t\to\infty}\frac{\ln t}t=0 $$ (this last one, you can do by L'Hôpital if you need to).