Could someone please help me to solve this limit. I have tried in Wolfram Alpha and it gave me the answer of 6.
But could someone please give me a step by step derivation please?
$$\lim_{x\to 0^+ }{\frac{x^2}{\ln\Bigl[\int_0^{x^{\frac{2}{3}}}{e^\frac{t^2}{2}dt}+1-x^{\frac{2}{3}}\Bigr]}}$$
Thanks
On
$$ \int_0^{x^{\frac{2}{3}}}{e^\frac{t^2}{2}dt}=\int_0^{x^{\frac{2}{3}}} {[1+\frac{t^2}{2}+o(t^2)]dt}=x^{2/3}+\frac{x^2}{6}+o(x^2)\\ \Rightarrow \ln[{\int_0^{x^{\frac{2}{3}}}{e^\frac{t^2}{2}dt}+1-x^{\frac{2}{3}}}]= \ln[1+\frac{x^2}{6}+o(x^2)]=\frac{x^2}{6}+o(x^2)\\ \Rightarrow \frac{x^2}{\ln[{\int_0^{x^{\frac{2}{3}}}{e^\frac{t^2}{2}dt}+1-x^{\frac{2}{3}}}]}=\frac{x^2}{\frac{x^2}{6}+o(x^2)}=6. $$
It helps to start by simplifying the expression. Letting $x^{2/3}=u$, we get
$${x^2\over\ln\left(\int_0^{x^{2/3}}e^{t^2/2}dt+1-x^{2/3} \right)}={u^3\over\ln\left(\int_0^ue^{t^2/2}dt+1-u \right)}$$
with $u\to0^+$ as the limit. L'Hopital is now relatively straightforward:
$$\begin{align} \lim_{u\to0^+}{u^3\over\ln\left(\int_0^ue^{t^2/2}dt+1-u \right)} &=\lim_{u\to0^+}\left({3u^2\over e^{u^2/2}-1}\left(\int_0^ue^{t^2/2}dt+1-u \right)\right)\\ &=\lim_{u\to0^+}{3u^2\over e^{u^2/2}-1}\quad(\text{since }\int_0^0e^{t^2/2}dt+1-0=0+1-0=1)\\ &=\lim_{v\to0^+}{6v\over e^v-1}\quad(\text{letting }v=u^2/2)\\ &=6\quad\text{(by L'Hopital again, if you like)} \end{align}$$
The real take-home lesson here is that since $x$ is a variable being taken to $0$, there is really no need to have fractional powers of it in the expression. A simple rewriting can make life ever so much easier.