<s>[INST] Solution verification: $\lim_{x\to 0}\frac{e^{2\sin x}\cosh x-1}{\sqrt[3]{x(\cosh x-\cos x)}}$ 
Find (without L'Hospital):$$\lim_{x\to 0}\frac{e^{2\sin x}\cosh x-1}{\sqrt[3]{x(\cosh x-\cos
 x)}}$$

My attempt:
$$\lim_{x\to 0}\cosh x=1$$
$$\lim_{x\to 0}e^{2\sin x}\cosh x-1=\lim_{x\to 0}\frac{e^{2\sin x}-1}{2\sin x}\cdot2\sin x=2\lim_{x\to 0}\sin x$$
$$\lim_{x\to 0}(\cosh x-\cos x)=\lim_{x\to 0}(1-\cos x)=\lim_{x\to 0}\frac{1-\cos x}{x^2}\cdot x^2=\frac{1}{2}\lim_{x\to 0}x^2$$
$$\lim_{x\to 0}\sqrt[3]{x(\cosh x-\cos x)}=\lim_{x\to 0}\sqrt[3]{x\cdot\frac{1}{2}x^2}=\frac{1}{\sqrt[3]{2}}\lim_{x\to 0}x$$
$$2\sqrt[3]{2}\lim_{x\to 0}\frac{\sin x}{x}=2\sqrt[3]{2}$$
Is this legitimate?
 [/INST] $$L=\lim_{x \rightarrow 0}\frac{e^{2 \sin x} \cosh x-1}{\sqrt[3]{x(\cosh x-\cos x)}}= \lim_{x\rightarrow 0} \frac{(1+2x+O(x^2))(1+x^2/2+O(x^4))-1}{{\sqrt[3]{x^3+O(x^5)}}}$$ $$\implies L= \lim_{x\rightarrow 0} \frac{2x+x^2/2+x^3+O(x^4)}{x}=2.$$
Here we have used $\cosh x \approx 1+x^2/2 +O(x^4), \cos x \approx 1-x^2/2+O(x^4), \sin x \approx x|O(x^3),$ $e^{x}\approx 1+x+O(x^2),$ when $x \rightarrow 0$
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