Evaluate $lim _{xto 0} frac{e^x+e^{-x}-2}{xtan x}$

Taylor series help a lot.

$$y= frac{e^x+e^{-x}-2}{xtan (x)}$$ $$y=frac{left(1+x+frac{x^2}{2}+frac{x^3}{6}+frac{x^4}{24}+Oleft(x^5right) right)+left(1-x+frac{x^2}{2}-frac{x^3}{6}+frac{x^4}{24}+Oleft(x^5right) right)-2 } {x left(x+frac{x^3}{3}+frac{2 x^5}{15}+Oleft(x^6right) right)}$$ Simplify and use long division to get $$y=1-frac{x^2}{4}+Oleft(x^3right)$$ which shows the limit and also how it is approached.

$$lim_{xto 0} frac{e^x+e^{-x}-2}{xtan x}=lim_{xto 0}frac{(e^{2x}-2e^{x}+1)}{xe^x tan x}$$ $$=lim_{xto 0} frac{(e^{x}-1)^2}{xe^x tan x}$$ $$=lim_{xto 0} left(frac{e^{x}-1}{x}right)^2frac{x}{tan x}frac{1}{e^x}$$ $$=1^2cdot 1cdot 1=color{blue}{1}$$

Hint:

$$dfrac{e^x+e^{-x}-2}x=dfrac{e^x-1+e^{-x}-1}x=dfrac{e^x-1}x-dfrac{e^{-x}-1}{-x}$$

$$implieslim_{xto0}dfrac{e^x+e^{-x}-2}{xtan x}=dfrac{lim_{xto0}dfrac{e^x+e^{-x}-2}x}{lim_{xto0}tan x}=dfrac00$$ right?

Better use

$$dfrac{e^x+e^{-x}-2}{xtan x}=left(dfrac{e^x-1}xright)^2dfrac x{sin x}cdotdfrac{cos x}{e^x}$$

Your mistake: $(frac {e^{x}-1}x -frac {e^{-x}-1} {-x}) to 0$ so you cannot just omit this factor.

By L'Hopital's Rule we get $lim frac {e^{x}-e^{-x}} {xsec^{2} x+tan x}$ and another application of L'Hopital's Rule gives $lim frac {e^{x}+e^{-x}} {2xsec^{2}x tan x+2sec ^{2} x}=1$