<s>[INST] How to find the value of $\lim_\limits{x \to 0}\frac{e^{x\cos{x^2}}-e^{x}}{x^5}$ How to find the exact value of the following limit?
$\lim_\limits{x \to 0}\frac{e^{x\cos{x^2}}-e^{x}}{x^5}$
 [/INST] HINT:
$$\dfrac{e^{x\cos x^2}-e^x}{x^5}=e^x\cdot\dfrac{e^{x(\cos x^2-1)}-1}{x^5}$$
$$\dfrac{e^{x(\cos x^2-1)}-1}{x^5}=\dfrac{e^{x(\cos x^2-1)}-1}{x(\cos x^2-1)}\cdot\dfrac{x(\cos x^2-1)}{x^5}$$
$$\dfrac{\cos x^2-1}{x^4}=-\left(\dfrac{\sin x^2}{x^2}\right)^2\cdot\dfrac1{(1+\cos x^2)}$$
Now use $\lim_{h\to0}\dfrac{e^h-1}h=1$  and $\lim_{y\to0}\dfrac{\sin y}y=1$
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