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?

Simply define $f(x) = e^x + e^{-x}.$ Then the expression equals
$$\frac{f(x) - f(0)}{x-0}\cdot \frac{x}{x^2+2x}.$$
As $x\to0,$ the first fraction $\to f'(0)$ by the definition of the derivative. The second fraction $\to 1/2.$ Since $f'(0)=0,$ the limit in question is $0\cdot (1/2) = 0.$
