$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}$ evaluation using expansion series $$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}$$
This is the limit which I got from the book of Joseph Edwards' Differential Calculus for Beginners the second exercise of the First Chapter Page -10 question number 19. Whose Answer was given as $\frac 1 2$.
So my try was as follows:
$$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}$$
$$\Rightarrow \lim \limits _{x \rightarrow 0} [ \frac{x^3(1 + \frac{x^4}{4} + \frac{x^8}{32}...) - (x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!}...)^{\frac 3 2} }{x^7} ]$$
Now, this is the quite a big problem, as we can see that sine expansion series has the power $\frac 3 2$ which is nearly impossible to evaluate.
The expansion series given in the book are: $(1+x)^n$ and $(1-x)^{-n}$ which are not useful here as there are infinite series where those are 2-terms.
Please anyone help me , I've just started Calculus :P
 A: You are on the right track. By using little-o notation you may control the expansions in a better way. We have that as $x\to 0^+$,
$$\frac{x^3(1 + \frac{x^4}{4} + o(x^4)) - (x^2 - \frac{x^6}{3!} + o(x^6))^{\frac 3 2} }{x^7}=\frac{1 + \frac{x^4}{4} + o(x^4) - (1 - \frac{x^4}{3!} + o(x^4))^{\frac 3 2} }{x^4}$$
Now all you need is that $(1+t)^{3/2}=1+\frac{3t}{2}+o(t)$ this can be shown by verifying that, 
$$\lim_{t\to 0}\frac{(1+t)^{3/2}-1}{t}=
\lim_{t\to 0}\frac{(1+t)^{3}-1}{t((1+t)^{3/2}+1)}=\lim_{t\to 0}\frac{3+o(1)}{((1+t)^{3/2}+1}=\frac{3}{2}.$$
A: We have $$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}{=\lim \limits _{x \rightarrow 0} \frac{e^{\frac {x^4}{4}} - \left({\sin x^2\over x^2}\right)^{3\over 2} }{x^4}\\=\lim \limits _{x \rightarrow 0} \frac{e^{\frac {x^4}{4}} -1+1- \left({\sin x^2\over x^2}\right)^{3\over 2} }{x^4}\\=\lim \limits _{x \rightarrow 0} \frac{e^{\frac {x^4}{4}} -1 }{x^4}+\lim \limits _{x \rightarrow 0} \frac{1 - \left({\sin x^2\over x^2}\right)^{3\over 2} }{x^4}}$$the latter equality is true since $\lim \limits _{x \rightarrow 0} \frac{e^{\frac {x^4}{4}} -1 }{x^4}={1\over 4}$ is bounded. It remains to show that $$\lim \limits _{x \rightarrow 0} \frac{1 - \left({\sin x^2\over x^2}\right)^{3\over 2} }{x^4}={1\over 4}$$This is  straight forward since $$\lim \limits _{x \rightarrow 0} \frac{1 - \left({\sin x^2\over x^2}\right)^{3\over 2} }{x^4}{=\lim \limits _{u \rightarrow 0} \frac{1 - \left({\sin u\over u}\right)^{3\over 2} }{u^2}\\=\lim \limits _{u \rightarrow 0} \frac{1 - \left({\sin u\over u}\right)^{3\over 2} }{u^2}{1 + \left({\sin u\over u}\right)^{3\over 2} \over 1 + \left({\sin u\over u}\right)^{3\over 2} }\\={1\over 2}\lim \limits _{u \rightarrow 0} \frac{1 - \left({\sin u\over u}\right)^3 }{u^2}\\={1\over 2}\lim \limits _{u \rightarrow 0} \frac{u^3-\sin^3 u }{u^5}\\={1\over 2}\lim \limits _{u \rightarrow 0} \frac{u^3-\left(u-{u^3\over 6}\right)^3 }{u^5}\\={1\over 2}\lim \limits _{u \rightarrow 0} \frac{1-\left(1-{u^2\over 6}\right)^3 }{u^2}\\={1\over 2}\lim \limits _{u \rightarrow 0} \frac{{u^2\over 6}\left(1+\left(1-{u^2\over 6}\right)+\left(1-{u^2\over 6}\right)^2\right) }{u^2}\\={1\over 2}\lim_{u\to 0}{{3u^2\over 6}\over u^2}\\={1\over 4}}$$hence the result.
