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.
Best Answer
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.