Two-variable limit $\frac{1-e^{(2x+y)^2}}{\sin^2(2x+y)}$

Question

How do I solve a limit for two variables

$$\lim_{(x,y)→(-1,2)} \frac{1-e^{(2x+y)^2}}{\sin^2(2x+y)}$$

Thanks.

Answer 1

It's really only a one-variable limit, because your function only depends on $2x+y$. Writing $2x+y=t$, and noting that $t \to 0$ when $x \to -1$ and $y \to 2$, $$ \lim_{(x,y) \to (-1,2)} \dfrac{1-e^{(2x+y)^2}}{\sin^2(2x+y)} = \lim_{t \to 0} \dfrac{1-e^{t^2}}{\sin^2(t)}$$

Answer 2

Hint: Use the facts $$\lim _{ x\rightarrow 0 }{ \frac { \sin { \left( x \right) } }{ x } } =1\\ \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } =1 } $$

Answer 3

Let $2x+y=u$ then the limit becomes,

$$\lim_{u \to 0} \frac{1-e^{u^2}}{\sin^2 u}$$

By Taylor series this becomes

$$\lim_{u \to 0} \frac{-u^2-\frac{1}{2}u^4+...}{\sin^2 u}$$

Given that,

$$\lim_{u \to 0} \frac{u}{\sin u}=1$$

The limit goes to $-1$ by the product and sum rules for limits.