Lim x→0(e^x^2-cos x-tfrac32x^2/(x^4 equals

Both numerator terms must be expanded to order $x^4$. $e^{x^2}=1+x^2+\dfrac{x^4}{2}+\cdots$ and $\cos x=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\cdots$. Then $e^{x^2}-\cos x=\dfrac32x^2+\left(\dfrac12-\dfrac1{24}\right)x^4+\cdots=\dfrac32x^2+\dfrac{11}{24}x^4+\cdots$. Subtracting $\tfrac32x^2$ leaves $\dfrac{11}{24}x^4+O(x^6)$, so the limit is $\dfrac{11}{24}$. L'Hopital would need four differentiations; the series route is faster. Forgetting the sign of the $\cos x$ term gives $\tfrac{13}{24}$. The same result follows from applying L'Hopital's rule four times, but the series method avoids long derivatives and shows exactly which terms survive, so it is the standard shortcut for such limits in JEE Main. Always keep terms up to the order of the denominator.