Solve $\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x}$ without using L'Hopital's rule I tried:
$$\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x} = \\
\frac{e^{x}(1+e^{-2x}-\frac{2}{e^x})}{x(x-2)} = \frac{e^x(1+e^{-2x})-2}{x(x-2)} = \frac{e^x(1+e^{-2x})}{x(x-2)}  - \frac{2}{x(x-2)} = \\
\frac{1+e^{-2x}}{x} \cdot  \frac{e^x}{x-2} - \frac{2}{x(x-2)} = ???$$
What do I do next?

Aside from not carelessly turning $x(x+2)$ into $x(x-2)$, partial fractions is the wrong tool for this job.  If you are allowed to used Taylor series, you can see that 
$$
e^x + e^{-x} - 2 = x^2 + O(x^x)$$
so you get the expression $$\frac{x^2}{x(x+2)} = \frac{x}{x+2} \to 0 $$
