Checking answer here to confirm if it's question mistake or mine: $\lim_{x \rightarrow 0} \frac{\tan (x - x)}{\sin (x-x)}$ = indetermined? why wrong? 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..?
 A: 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:
$$\lim_{x\to0}\frac{\tan x-x}{\sin x-x}=\lim_{x\to0}\frac{\sec^2x-1}{\cos x-1}=\lim_{x\to0}\frac{2\sec^2x\tan x}{-\sin x}=\lim_{x\to0}-2\sec^3x=-2.$$
A: They expect you to divide out the $\sin(x-x)$, get $\frac 1{\cos (x-x)}$ and say the limit is $1$.  They are lazily making use of the fact that we can cancel terms that look like $0$ in limits because we do not evaluate them when $x$ is at the limit.
They are missing the fact that you substitute the same value of $x$ everywhere it appears in the expression.  When you do that, you get $\frac 00$ uniformly, regardless of the value of $x$.  Even when $x \neq 0$ you cannot calculate the value of the expression.  I agree with you that the answer should be undefined.
A: As noticed the given expression for the limit is meaningless.
Another possible typo could be for:
$$\lim_{x \rightarrow x_0} \frac{\tan (x - x_0)}{\sin (x-x_0)}=1$$
which is one of the option.
