# Limit of exponential function without L'hopital rule

Can anyone give me a hint on how to solve the following expression?

Solve for m

$\lim_{x \to 0} f(x) = \frac{(e^x+e^-x)(sin(mx)}{(e^x-1)}$

given that

$\lim_{x \to 0} f(x) =4 + m$

I know that it is $\frac{0}{0}$ form. I tried to do this approach but it does not seem right.

**L'hopital rule would help alot but it is not allowed in my class. **

$\lim_{x \to 0} f(x) = \frac{(e^x+e^-x)(sin(mx)}{(e^x-1)}\cdot \frac{e^x+1}{e^x+1}$

Using Wolfram Alpha, the answer for m should be m = 4.

• @JensSchwaiger My expression is correctly typed. Commented Jul 4, 2018 at 3:27
• @WolverIng it isn't. Commented Jul 4, 2018 at 3:32
• @WolverIng It's not well-parenthesized. Commented Jul 4, 2018 at 3:33
• @WolverIng Hint: write it as $\;m(e^x+e^{-x}) \cdot \frac{x}{e^x-1} \cdot \frac{\sin(mx)}{mx}\,$.
– dxiv
Commented Jul 4, 2018 at 3:36

Supposing that $f$ is given as $f(x)=\frac{(e^x+e^{-x})(\sin(mx))}{(e^x-1)}$ and that $\lim_{x\to0}f(x)=4+m$ multiplying the expression for $f$ with $m \frac{x}{mx}$ would probably be helpful.
The limit of $$f(x)=\frac{\left(e^x+e^{-x}\right) \sin (m x)}{e^x-1}$$ can "easily" be obtained using the following equivalents
$$e^x \sim 1+x \qquad e^{-x}\sim 1-x \qquad \sin(mx) \sim mx \implies f(x)\sim 2m$$
You could go further using Taylor expansions for each piece $$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O\left(x^4\right)$$ $$e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+O\left(x^4\right)$$ $$\sin(mx)=m x-\frac{m^3 x^3}{6}+O\left(x^4\right)$$ which make $$f(x)=\frac{ 2 m x+\left(m-\frac{m^3}{3}\right) x^3+O\left(x^4\right)}{x+\frac{x^2}{2}+\frac{x^3}{6}+O\left(x^4\right) }$$ Now, long division to get $$f(x)=2 m-m x+\frac{1}{6} m \left(7-2 m^2\right) x^2+O\left(x^3\right)$$ which shows the limit and how it is approached.