I need help taking the limit of this problem please:
f(x) = sin(x) / [x + tan(x)] as x approaches 0
Any help would be appreciated, thank you.
Thankfully the edit you just did has removed a confusing ambiguity.
The lazy way is to use l'Hopitals Rule. Have you been taught it?
Alternatively, you can substitute the Maclaurin Series for sin x and tan x, then cancel out the common factor of x in the numerator and denominator and then take the limit without trouble.
In that case:
$\displaystyle \lim_{x\to0}\frac{\sin x}{x+\tan x}=\lim_{x\to0}\frac{\sin x}{x[1+(\tan x)/x]}=\lim_{x\to0}\left(\frac{\sin x}x\right)\left(\frac1{1+(\tan x)/x}\right)$
$\displaystyle =\lim_{x\to0}\left(\frac{\sin x}x\right)\lim_{x\to0}\frac1{1+(\sin x)/(x\cos x)}$
$\displaystyle =1\cdot\lim_{x\to0}\frac1{1+[(\sin x)/x][1/\cos x]}$
$\displaystyle =\frac1{1+[\lim_{x\to0}(\sin x)/x][\lim_{x\to0}(1/\cos x)]}$
$\displaystyle =\frac1{1+1\cdot1} = \frac12$