I encountered a questions with asking to evaluate: $\lim_{x \rightarrow 0} \frac{\tan (x – x)}{\sin (x-x)}$, and gives multiple choice options $-2, 0, 1, 2$ but doesn't have indetermined…
If I used trig identities, I'd get $\frac{1}{\cos (x-x)}$ by converting tan to $\frac {\sin}{\cos}$ so the best guess is 1, but that's still wrong.
Please can someone tell me where my gap is? I'm learning l'Hopital rule if that helps, but even applying that gives $\frac{0}{0} = \text{indetermined}.$ Could the question be wrong..?
Best Answer
Are you sure those parentheses were there? Maybe it was a typo. Observe that $$\lim_{x\to0}\frac{\tan x-x}{\sin x-x}=-2$$ which was one of the choices. Without those silly parentheses, $$\sin x-x=(\sin x)-x=-\frac16x^3+\frac1{120}x^5-\cdots$$ and $$\tan x-x=(\tan x)-x=\frac13x^3+\frac2{15}x^5+\cdots.$$
The limit of the quotient can be seen immediately from the Maclaurin series. Somewhat more tediously, we can apply l'Hospital's rule twice: $$\lim_{x\to0}\frac{\tan x-x}{\sin x-x}=\lim_{x\to0}\frac{\sec^2x-1}{\cos x-1}=\lim_{x\to0}\frac{2\sec^2x\tan x}{-\sin x}=\lim_{x\to0}-2\sec^3x=-2.$$