How to solve the limit $\lim_{(x,y)\to(2,3)}\frac{\sin(x)+2}{(x^2-y^2+5)^2}$

Question

How can I solve the following limit or prove that the limit does not exist in $\Bbb R$:

$$ \lim_{(x,y)\to(2,3)}\frac{\sin(x)+2}{(x^2-y^2+5)^2}? $$

I tried to prove the limit does not exist with $y=3x/2$, $y=x+1$, but it looks like a dead end.

Answer 1

Note that since $-1 \le \sin x \le 1$, we have: $$\frac{1}{\left(x^2-y^2+5\right)^2} \le \frac{\sin x +2}{\left(x^2-y^2+5\right)^2} \le \frac{3}{\left(x^2-y^2+5\right)^2}$$ And the denominator clearly tends to $0$ (but stays positive) when $(x,y) \to (2,3)$, so...

Answer 2

Hint: The denominator goes always (no matter how you choose $y$ and $x$) to $0$ and the numerator goes to a number different than $0$, since $\sin(x)\in [-1,1]$.

Answer 3

Note that

• $\sin(x)+2\to\sin 2 +2>0$
• $(x^2-y^2+5)^2\to0^+$