evaluation of limits of non-elementary functions

Question

I want to try and evaluate the following limit: $$L_1=\lim_{x\to 0}\frac{\text{erf}(x^2)}{\text{erf}(x)}.$$ If I use L'Hopital's rule and then Leibniz' integral rule, I believe I get $$L_1=\lim_{x\to 0}\frac{d[\text{erf}(x^2)]}{d[\text{erf}(x)]},$$ and I believe $$\frac{d[\text{erf}(x^2)]}{dx}=\frac{d}{dx}\left[\frac{2}{\sqrt{\pi}}\int_0^{x^2}e^{-t^2}dt\right]=\frac{4}{\sqrt{\pi}}xe^{-x^4}$$ $$\frac{d[\text{erf}(x)}{dx}=\frac{2}{\sqrt{\pi}}e^{-x^2}$$ and so $$L_1=\lim_{x\to 0}\frac{2xe^{-x^4}}{e^{-x^2}}$$ since $e^{-x^2}$ and $e^{-x^4}$ evaluate to $1$, we can assume $L_1=0.$

Is this correct?

Answer 1

There is no need of L'Hospital's Rule here. By fundamental theorem of calculus we have $$\lim_{x\to 0}\frac{\operatorname {erf} (x)} {x} =\frac{2}{\sqrt{\pi}}\lim_{x\to 0}\frac{1}{x}\int_{0}^{x}e^{-t^2}\,dt=\frac{2}{\sqrt{\pi}} e^{-0^2}=\frac{2}{\sqrt {\pi}} $$ and then $$\lim_{x\to 0}\frac{\operatorname {erf} (x^2)}{\operatorname {erf} (x)} =\lim_{x\to 0}\frac{\operatorname {erf}(x^2)}{x^2}\cdot\frac{x}{\operatorname {erf} (x)} \cdot x=\frac{2}{\sqrt{\pi}}\cdot\frac{\sqrt{\pi}}{2}\cdot 0=0$$

Answer 2

Another way could be Taylor expansion to get the limit and even more.

Since, around $t=0$, $$\text{erf}(t)=\frac{2 t}{\sqrt{\pi }}-\frac{2 t^3}{3 \sqrt{\pi }}+O\left(t^5\right)$$ $$\frac{\text{erf}(x^2)}{\text{erf}(x)}=\frac{\frac{2 x^2}{\sqrt{\pi }}-\frac{2 x^6}{3 \sqrt{\pi }}+O\left(x^8\right) } {\frac{2 x}{\sqrt{\pi }}-\frac{2 x^3}{3 \sqrt{\pi }}+O\left(x^5\right) }=x+\frac{x^3}{3}+O\left(x^5\right)$$ which shows the limit and how it is approached.