Baffled with $\lim\limits_{x\to 0^+}{1+\ln^2 x\over x}$

Question

Solve

I)$$\lim\limits_{x\to+\infty}{f(x)}=\lim\limits_{x\to +\infty}{{1+\ln^2x\over x}}$$

II)$$\lim\limits_{x\to 0^+}{f(x)}=\lim\limits_{x\to 0^+}{{1+\ln^2x\over x}}$$

Personal work:

I)$$\lim\limits_{x\to +\infty}{{1+\ln^2x\over x}}=\lim\limits_{x\to +\infty}({1\over x}+{\ln^2x\over x})=L$$

Let $u=\ln x \iff e^u=x$, hence $u_0=+\infty$, then: $$\lim\limits_{u\to +\infty}({1\over e^u}+{u^2\over e^u})=\lim\limits_{u\to +\infty}{1\over e^u}+\lim\limits_{u\to+\infty}{u^2\over e^u}=\cdots=0+0^+=0$$ because, when $${1\over e^u}\to +\infty$$, then $${1\over e^u}=0^+.$$ So, $$L=0.$$

II) I'm a bit baffled about letting either $u=\ln x$ or $u=x.$

Answer 1

Note that it is not an indeterminate form thus

$$\lim\limits_{x\to 0^+}{f(x)}=\lim\limits_{x\to 0^+}{{1+\ln^2x\over x}}=\left(\frac{+\infty}{0^+}\right)=+\infty$$

or also

$$\lim\limits_{x\to 0^+}{f(x)}=\lim\limits_{x\to 0^+} \frac1x\cdot(1+\ln^2x)=\left(+\infty\cdot+\infty\right)=+\infty$$

Answer 2

For the first one: $$\lim_{x \to +\infty} \frac{1+ln^2x}{x}=\lim_{x \to +\infty} \frac{1}{x} + \frac{ln^2x}{x} $$ where $$\lim_{x \to +\infty}\frac{1}{x}=0$$ and $$\lim_{x \to +\infty}\frac{ln^2x}{x}=(By L'Hospital's rule)=\lim_{x \to +\infty} 2\frac{lnx}{x}=(L'Hospital)=\lim_{x \to +\infty}2\frac{1}{x}=0$$ So: $$\lim_{x \to +\infty} \frac{1+ln^2x}{x}=0+0=0$$

If you don't know about L'Hospital's rule, it's a way of finding limits of indeterminate form using derivatives.

For the second one: $$\lim_{x \to 0^+}\frac{1+ln^2x}{x}= \lim_{x \to 0^+}(1+ln^2x)\frac{1}{x}=(+\infty)(+\infty)=+\infty$$