Finding $\lim\limits_{x\to0}x^2\ln (x)$ without L'Hospital

Solution 1:

Note that we should rather consider $$\lim_{x\to 0^+}x^2\ln x.$$ Substitute $x$ with $e^{-t}$ (this is possible for $x>0$) to get $$ \lim_{x\to 0^+}x^2\ln x=\lim_{t\to+\infty}(-t)(e^{-t})^2=\lim_{t\to+\infty}\frac{-t}{(e^t)^2}$$ and use our favorite estimate for the exponential function: $e^t\ge 1+t$, to get $$ \left|\lim_{x\to 0^+}x^2\ln x\right|\le \lim_{t\to+\infty}\left|\frac{-t}{(e^t)^2}\right|=\lim_{t\to+\infty}\frac{t}{(e^t)^2}\le \lim_{t\to+\infty}\frac{t}{(t+1)^2}=0.$$

Solution 2:

Sandwich theorem, using the fact (as far as I know it is new)

$$ |\ln(x)| < \frac{1}{x}\,, $$

when $x$ close to $0$, we have

$$ |x^2\ln (x)| < x. $$

Solution 3:

Just to use FTC: for $0<x\leq 1$$$ 0<-x\ln x=x\int_x^1\frac{1}{t}dt=\int_x^1\frac{x}{t}dt\leq \int_x^11dt=1-x\leq 1\Rightarrow 0<-x^2\ln x\leq x. $$ Hmm... that's Mhenni Benghorbal's argument. But since I prove the inequality, I guess I'll leave it.

Solution 4:

By mean value theorem there's $\xi_x\in (x,1)$ such that $$\log(x)-\log(1)=(x-1)\frac{1}{\xi_x}$$ and since $$x\leq\frac{x}{\xi_x}\leq1$$ we have $$x(x-1)\leq x^2 \log x = x^2 (x-1)\frac{1}{\xi_x}\leq x^2(x-1)$$ so we conclude by squeeze theorem.

Solution 5:

For an elementary way consider this:

$\lim_{x\to 0} x^2 \ln x$

$\lim_{x\to 0} \ln {e^{x^2}} \ln x$

$\lim_{x\to 0} \ln [(e^{x^2})^{\ln x}]$

$\lim_{x\to 0} \ln[(e^{\ln x})^{x^2}]$

$\lim_{x\to 0} \ln x^{x^2}$

$\lim_{x\to 0} \ln (x^x)^x$

It is a well-known fact that $\lim_{x\to 0} x^x=1$ , so if we substitute that in we get $\lim_{x\to 0} \ln 1^x$ , or $\lim _{x\to 0} \ln 1=0$ . So our original limit is equal to $0$ . Let me know if anything is not clear. Hope this helps.