Existence of Multivariable limit with sin

Question

Im trying to solve this limit $$ \lim_{(x,y)\to(2,2)} \frac{\sin(4-xy)}{16-x^2y^2} $$

For a reason Wolfram cant compute it (maybe I'm using it wrong) but anyways I saw that someone solve it using $t=xy$ and so the limit would be

$$ \lim_{t\to 4} \frac{\sin(4-t)}{16-t^2}=\frac 18 $$

But does this prove limit exist or this is only for one path?

Answer 1

Your reasoning is perfectly fine and correct, indeed for any path

$$t=4-xy \to 0 \implies \frac{\sin(4-xy)}{4-xy}=\frac{\sin t}{t} \to 1$$

and since for any path $4+xy \to 8$ we have

$$\frac{\sin(4-xy)}{16-x^2y^2}=\frac{\sin(4-xy)}{4-xy} \frac{1}{4+xy}\to 1 \cdot \frac18=\frac18$$

---

The following are some examples of similar limits with discussion about path dependence

• Why doesn't $ \lim_{(x,y)\rightarrow(0,0)} xy\sin(\frac{1}{xy}) $ exist?
• Limit $\lim\limits_{(x,y)\to(0,0)}\frac{\sin(x^2-y^2)}{x^2-y^2}$
• Limit $\lim_{(x,y)\to (0,0)}\frac{\sin(xy)}{xy}$
• Two-variable limit of $\lim_{(x,y)\to(0,0)}\frac{\sin(x^4+y^4)}{x^2+y^2}$
• The limit of a two-variable function $\lim_{ (x, y) \to (0,0) } \frac{ xy \sin y }{ x^2 + y^2 }$
• How to find $\lim_{(x,y)\rightarrow(0,0)} \frac{x^2\sin^2(y)}{x^2+3y^2}$

Answer 2

You can write your limit as $$ \lim_{(x,y)\to(2,2)}\dfrac{\sin(4-xy)}{16-x^2y^2} = \left(\lim_{(x,y)\to(2,2)} \frac{1}{4+xy}\right)\cdot\left(\lim_{(x,y)\to(2,2)} \dfrac{\sin(4-xy)}{4-xy}\right) = \frac 18 \lim_{t \to 0} \frac{\sin t}{t} = \frac 18 $$

the calculation of the second factor is justified by the limit of a composition.