로피탈 정리

The first thing we notice is, regardless of $k$, when we plug in $x=0$ both numerator and denominator become zero. Let's use this fact, and apply l'Hopital's Rule.

$$\begin{aligned} \lim\limits_{x\rightarrow 0}\underbrace{\dfrac{e^{k\sin(x^2)}-(1+2x^2)}{x^4}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to0}} &=\lim_{x\rightarrow 0}\dfrac{2kx\cos(x^2)e^{k\sin(x^2)}-4x}{4x^3}\\ &=\lim_{x\rightarrow 0}\dfrac{2k\cos(x^2)e^{k\sin(x^2)}-4}{4x^2} \\ \text{When we plug in $x=0$, the denominator becomes 0, and the numerator becomes $2k-4$. So, we'll need some cases, because the behaviour of the limit depends on $k$.} \\ \\ \text{For $k=2$:} \\ \lim_{x\rightarrow 0}\dfrac{2k\cos(x^2)e^{k\sin(x^2)}-4}{4x^2}&= \lim_{x\rightarrow 0}\underbrace{\dfrac{4\cos(x^2)e^{2\sin(x^2)}-4}{4x^2}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to0}}\\ &= \lim_{x\rightarrow 0}\dfrac{-8x\sin(x^2)e^{2\sin(x^2)} +16x\cos^2(x^2)e^{2\sin(x^2)}}{8x}\\ &= \lim_{x\rightarrow 0}\big[-\sin(x^2)e^{2\sin(x^2)} +2\cos^2(x^2)e^{2\sin(x^2)}\big]\\ &=2 \\ \text{For $k>2$, the numerator goes to $2k-4$, which is a positive constant, while the denominator goes to $0$ from the right, so:} \\ \lim_{x\rightarrow 0}\dfrac{2k\cos(x^2)e^{k\sin(x^2)}-4}{4x^2}&= \infty \\ \text{For $k<2$, the numerator goes to $2k-4$, which is a negative constant, while the denominator goes to $0$ from the right, so:} \\ \lim_{x\rightarrow 0}\dfrac{2k\cos(x^2)e^{k\sin(x^2)}-4}{4x^2}&= -\infty \end{aligned}$$