Solve $\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x}$ without using L'Hopital's rule

Question

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?

Answer 1

The path you took leads nowhere, because you get a $\infty-\infty$ form. (By the way, $x^2+2x=x(x+2)$, not $x(x-2)$, but it's a detail.)

Use the fact that $$ e^x+e^{-x}-2=e^{-x}(e^x-1)^2 $$ so your limit can be written as $$ \lim_{x\to0}\frac{e^{-x}}{x+2}\frac{e^x-1}{x}(e^x-1) $$

Of course, l’Hôpital in this case is much easier:

$$\displaystyle\lim_{x\to0}\frac{e^x-e^{-x}}{2x+2}$$

Answer 2

You can use hyperbolics: $2\cosh x=e^x+e^{-x}$ and so the limit expression can be written as $$\frac{2}{x+2}\cdot\frac{\cosh x-1}{x-0}$$ and use the limit definition of the derivative on $y=\cosh x$ taken at $x=0$. Now give it a try from here.

Answer 3

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 $$\frac{x^2}{x(x+2)} = \frac{x}{x+2} \to 0 $$

Answer 4

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 $0\cdot (1/2) = 0.$