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$$lim limits _{x rightarrow 0} frac{x^3 e^{frac {x^4}{4}} - {sin^{frac 3 2} {x^2}} }{x^7}$$
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

You are on the right track. By using little-o notation you may control the expansions in a better way. We have that as $xto 0^+$,
$$frac{x^3(1 + frac{x^4}{4} + o(x^4)) - (x^2 - frac{x^6}{3!} + o(x^6))^{frac 3 2} }{x^7}=frac{1 + frac{x^4}{4} + o(x^4) - (1 - frac{x^4}{3!} + o(x^4))^{frac 3 2} }{x^4}$$
Now all you need is that $(1+t)^{3/2}=1+frac{3t}{2}+o(t)$ this can be shown by verifying that,
$$lim_{tto 0}frac{(1+t)^{3/2}-1}{t}=
lim_{tto 0}frac{(1+t)^{3}-1}{t((1+t)^{3/2}+1)}=lim_{tto 0}frac{3+o(1)}{((1+t)^{3/2}+1}=frac{3}{2}.$$

We have $$lim limits _{x rightarrow 0} frac{x^3 e^{frac {x^4}{4}} - {sin^{frac 3 2} {x^2}} }{x^7}{=lim limits _{x rightarrow 0} frac{e^{frac {x^4}{4}} - left({sin x^2over x^2}right)^{3over 2} }{x^4}\=lim limits _{x rightarrow 0} frac{e^{frac {x^4}{4}} -1+1- left({sin x^2over x^2}right)^{3over 2} }{x^4}\=lim limits _{x rightarrow 0} frac{e^{frac {x^4}{4}} -1 }{x^4}+lim limits _{x rightarrow 0} frac{1 - left({sin x^2over x^2}right)^{3over 2} }{x^4}}$$the latter equality is true since $lim limits _{x rightarrow 0} frac{e^{frac {x^4}{4}} -1 }{x^4}={1over 4}$ is bounded. It remains to show that $$lim limits _{x rightarrow 0} frac{1 - left({sin x^2over x^2}right)^{3over 2} }{x^4}={1over 4}$$This is straight forward since $$lim limits _{x rightarrow 0} frac{1 - left({sin x^2over x^2}right)^{3over 2} }{x^4}{=lim limits _{u rightarrow 0} frac{1 - left({sin uover u}right)^{3over 2} }{u^2}\=lim limits _{u rightarrow 0} frac{1 - left({sin uover u}right)^{3over 2} }{u^2}{1 + left({sin uover u}right)^{3over 2} over 1 + left({sin uover u}right)^{3over 2} }\={1over 2}lim limits _{u rightarrow 0} frac{1 - left({sin uover u}right)^3 }{u^2}\={1over 2}lim limits _{u rightarrow 0} frac{u^3-sin^3 u }{u^5}\={1over 2}lim limits _{u rightarrow 0} frac{u^3-left(u-{u^3over 6}right)^3 }{u^5}\={1over 2}lim limits _{u rightarrow 0} frac{1-left(1-{u^2over 6}right)^3 }{u^2}\={1over 2}lim limits _{u rightarrow 0} frac{{u^2over 6}left(1+left(1-{u^2over 6}right)+left(1-{u^2over 6}right)^2right) }{u^2}\={1over 2}lim_{uto 0}{{3u^2over 6}over u^2}\={1over 4}}$$hence the result.