limit of $\frac{x^2}{\ln{|x|}}$

Question

For unknown reasons I have more problems solving this limit than expected.

I want to determine the limit: $\lim_{x \to 0} \sqrt[3]{\dfrac{x^2}{\ln(|x|)}}$.

Using graphic plotter, the limit exists and has to be 0. Well, being rough $\frac{0}{\infty} = 0$. But that is not very mathematical. Also, L'Hospital does not work for this.

I tried using $t(x) = \frac{1}{x}$, which means $\lim_{t \to \infty} \sqrt[3]{\dfrac{1}{t^2\ln(|\frac{1}{t}|)}}$, maybe in order to use L'Hospital here. But nope....

Can someone help me? Really appreciate your help.

Answer 1

Since $\lim_{x \to 0} x^2 = 0$ and $\lim_{x \to 0} \log(\lvert x \rvert) = - \infty$, we conclude that $$ \lim_{x \to 0} \frac{x^2}{\log(\lvert x \rvert)} = 0. $$ Additionaly, because $\mathbb{R} \to \mathbb{R}, x \mapsto \sqrt[3]{x}$ is continuous, we deduce that $$ \lim_{x \to 0} \sqrt[3]{\frac{x^2}{\log{\lvert x \rvert}}} = 0. $$ In general, we have the following.

Proposition. Let $X$ be a non-empty subset of $\mathbb{R}$, let $f, g \colon X \to \mathbb{R}$ be two functions, and let $x_0$ be a limit point of $X$. If $\lim_{x \to x_0} f(x) = 0$ and $\lim_{x \to x_0} g(x) = + \infty$, then $$ \lim_{x \to x_0} \frac{f(x)}{g(x)} = 0. $$ Proof. Let $\varepsilon$ be an arbitrary positive real number. Since $\lim_{x \to x_0} g(x) = + \infty$, there exists some $\delta_1 > 0$ such that $\lvert g(x) \rvert > 1$, for all $x \in X$ with $0 < \lvert x - x_0 \rvert < \delta_1$. Moreover, given that $\lim_{x \to x_0} f(x) = 0$, there exits some $\delta_2 > 0$ such that $\lvert f(x) \rvert < \varepsilon$, for all $x \in X$ with $0 < \lvert x - x_0 \rvert < \delta_2$. Take $\delta := \min\{\delta_1,\delta_2\} > 0$ and let $x$ be an arbitrary point of $X$ such that $0 < \lvert x - x_0 \rvert < \delta$. On one hand, we have that $0 < \lvert x - x_0 \rvert < \delta_1$, which means that $\lvert g(x) \rvert > 1$. On the other hand, we know that $0 < \lvert x - x_0 \rvert < \delta_2$, which implies that $\lvert f(x) \rvert < \varepsilon$. Note that $\frac{1}{\lvert g(x) \rvert} < 1$ (and it makes sense to talk about $\frac{1}{\lvert g(x) \rvert}$ [why?]). Putting all this together we conclude $\left\lvert \frac{f(x)}{g(x)} \right\rvert = \frac{\lvert f(x) \rvert}{\lvert g(x) \rvert} < \varepsilon$. Therefore, $\lim_{x \to x_0} \frac{f(x)}{g(x)} = 0$. $\square$

An analogous argument allows you to prove this statement for the case $\lim_{x \to x_0} g(x) = - \infty$ or $\lim_{x \to x_0} f(x) = L$, for any real number $L$.

Answer 2

"Well, being rough $\frac 0∞=0$. But that is not very mathematical. "

Then make it mathematical.

We need to show that for every $\epsilon > 0$ there is a $\delta > 0$ where $-\delta < x < \delta; x\ne 0 \implies -\epsilon <\frac {x^2}{\ln |x|} < \epsilon$.

As we know $x^2 \to 0$ and $\ln |x|\to -\infty$ we can achieve that by choosing a $\delta$ so that $-\delta < x < \delta;x\ne 0\implies 0 < x^2 < \epsilon$ and $\ln|x| < -1$.

.... That's it. Do that.

If $0 < \epsilon$. We know $x^2 \to 0$ so there is $\delta_1 >0$ where $0< x < \delta_1$ implies $x^2 < \epsilon$. (We even know that $\delta_1 = \min (\sqrt \epsilon, 1)$ will do but we don't need to know that. That we know $x^2\to 0$ is enough to cite that such a $\delta_1$ exists.)

We know $\ln |x| \to -\infty$ so for $N=-1$ we we know there is a $\delta_2 > 0$ where $0< x < \delta_2$ implies $\ln|x| < N=-1$. (We even knew that $\delta_2 < \frac 1e$ will do but we don't need to know that. Again it's enough to just cite $\ln|x| \to -\infty$)

So let $\delta = \min(\delta_1, \delta_2)$ Then if $0 < x < \delta$ we get $-\epsilon < \frac {x^2}{\ln |x|}| < 0$.

So $\lim_{x\to 0^+}\frac{x^2}{\ln|x|} = 0$.

Answer 3

As noticed the limit is clearly $0$ since it is in the form $\frac 0 {\infty}$ which is not an indeterminate form.

Note that this is an intuitive fact but this is also a theorem in calculus

$$\lim_{x\to x_0} f(x)=L\in \mathbb R \;\land\;\lim_{x\to x_0} g(x)=\infty \implies \lim_{x\to x_0} \frac{f(x)}{g(x)}=0$$

therefore we can directly conclude from there that limit is equal to zero.

As an alternative, if you don't like logarithm, assuming wlog $x>0$, by $x=e^{-y}$ with $y\to \infty$ we obtain

$$\dfrac{x^2}{\ln(|x|)}=\frac{1}{-ye^{2y}}\to 0 \implies \sqrt[3]{\dfrac{x^2}{\ln(|x|)}} \to 0$$

(we are also assuming the definition for cube root extended to negative numbers)