How to compute $\lim_{x \to 0^+} 1 + \frac{\ln(x)}{x^2}$?

Question

How to compute the following limit?

$$\lim_{x \to 0^+} 1 + \frac{\ln(x)}{x^2}$$

I tried factoring $x^2$, $\ln(x)$, tried replacing $x^2$ with $e^{2\ln(x)}$ but nothing could remove the indeterminate forms.

Answer 1

First, we can get rid of the leading $1$ with limit laws: $\lim_{x \to 0^+} 1 + \frac{\ln x}{x^2} = 1 + \lim_{x \to 0^+} \frac{\ln x}{x^2}.$ Let $f(x) = \frac{\ln x}{x^2}.$

As I said in the comments the limit tends towards $-\infty,$ and what this would precisely mean is that for any real number $\delta,$ there is some real $\epsilon > 0$ such that $f(x) < \delta$ for all $x \in (0, \epsilon).$

The first step here is to show that $f(x)$ is always increasing for positive $x$ near $0$, which we can do by looking at the derivative: $$f'(x) = \frac{(x^2)(\frac1x) - (\ln x)(2x)}{x^4} = \frac{1 - 2\ln(x)}{x^3}$$

For all $0 < x < 1,$ the numerator and denominator will both be positive, so $f'$ will be positive. Because of this, we can say that for all $0 < x_1 < x_2 \leq 1,$ we have that $f(x_2) - f(x_1) = \int_{x_1}^{x_2} f'(t) dt > 0,$ so $f(x_1) < f(x_2).$

Now, consider that if $\delta > -e^2,$ then we can simply let $\epsilon = \frac1e$: $f(\frac1e) = -e^2,$ and as argued before, $0 < x < \frac1e$ now implies $f(x) < -e^2 < \delta.$

If $\delta < -e^2,$ consider $\epsilon = \sqrt{\frac1{|\delta|}}.$ Since $\delta < -e^2$ we have $-\frac1\delta = \frac1{|\delta|} < \frac1{e^2},$ so $\sqrt{\frac1{|\delta|}} = \epsilon < \frac1e$ by the monotonicity of the square root. Also by the monotonicity of $\ln$ this means that $\ln \epsilon < -1.$

This means that $$f(\epsilon) = \frac{\ln\epsilon}{\epsilon^2} < \frac{-1}{\epsilon^2} = \frac{-1}{\frac1{|\delta|}} = -|\delta| = \delta$$

and by our earlier argument, $0 < x < \epsilon$ implies that $f(x) < f(\epsilon) < \delta,$ which is exactly what we set out to prove.