I'm asked to check if there exists $\lim_{(x,y) \to (0,0)} f(x,y)$ for the following function:
$$
f(x,y) =
\begin{cases}
\displaystyle \frac{\sin(x) – \sin(y)}{\tan(x) – \tan(y)} & \text{ if } \tan x \neq \tan y\\
\cos^{3} (x) & \text{otherwise}
\end{cases}
$$
I have a guess that the limit exists and is equal to $1$, but I'm unable both to prove it or deny it.
The most natural manipulation (expanding $\tan$'s in the denominator and applying trigonometric identities) gives me for $\tan x \neq \tan y$ that $f(x,y) = \cos x \cos y(\sin(x) – \sin(y))/\sin(x-y)$, which doesn't help much. Any ideas or a good counterexample for my conjecture?
Best Answer
$f(x,y)=\frac{2\cos{\frac{x+y}{2}}\sin{\frac{x-y}{2}}\cos{x}\cos{y}}{\sin{(x-y)}}$, when we use the formula $\sin{x}-\sin{y}=2\cos{\frac{x+y}{2}}\sin{\frac{x-y}{2}}$. Now we need to calculate $\lim\limits_{x\to0}\frac{2\sin{\frac{x-y}{2}}}{\sin{(x-y)}}$, since all other factors tend do $1$. Now $t=x-y$ and $t\to 0$ should give you the wanted result.