<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] The key is to use laws of algebra of limits whose primary function is to help evaluate the limit of a complicated function which is composed of many simpler expressions connected by algebraic operations $+, -, \times, /$ given the limits of these simpler expressions. I have described these laws in a simple manner in this answer. 
You should have in memory a table of well known standard limits which allow us to deal with the simpler expressions mentioned in last paragraph.
In this question we need the following limits $$\lim_{x\to 0}\frac{e^x-1}{x}=1=\lim_{x\to 0}\frac{\sin x} {x} $$ and $$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2}=\lim_{x\to 0}\frac{\cosh x-1}{x^2}$$ First we deal with denominator. We can rewrite it as $$x\sqrt[3]{\frac{\cosh x-1}{x^2}+\frac{1-\cos x} {x^2}}$$ The expression under radical sign tends to $(1/2)+(1/2)=1$ and hence the denominator can be safely replaced by $x$.
Next we deal with numerator. It can be rewritten as $$e^{2\sin x} \cosh x-\cosh x +\cosh x - 1$$ or $$\cosh x (e^{2\sin x} - 1)+\cosh x - 1$$ Thus the desired limit is equal to the limit of $$\cosh x\cdot\frac{e^{2\sin x} - 1}{2\sin x} \cdot 2\cdot\frac{\sin x} {x} +x\cdot\frac{\cosh x - 1}{x^2}$$ which is $$1\cdot 1\cdot 2\cdot 1 +0\cdot\frac{1}{2}=2$$
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