Prove that $\lim _{(x,y) \to (0,2)} (1+xy)^\frac{2}{x^2+xy} = e^2$

Question

Prove that $$\lim _{(x,y) \to (0,2)} (1+xy)^{\frac{2}{x^2+xy}} = e^2.$$

Can someone suggest me the way to prove it? I tried but couldn't get through the $\frac{2}{x^2+xy}$.

Answer 1

$$\lim _{(x,y) \to (0,2)} (1+xy)^{\frac{2}{x^2+xy}} = \lim _{(x,y) \to (0,2)} e^{\frac{2}{x^2+xy} \ln (1+xy)}$$ $$\lim _{(x,y) \to (0,2)} {\frac{2}{x^2+xy} \ln (1+xy)}=\lim _{(x,y) \to (0,2)} \frac{2xy}{x(x+y)} \frac{ \ln (1+xy)}{xy}$$

Answer 2

Hint:

$$ (g(x))^{f(x)} = e^{f(x)\ \text{ln} \ g(x)} $$

This is a general strategy that you can try if you encounter difficult exponents when trying to take limits.

(in general you may need to check that the domain of the logarithm is being respected when using this trick - perhaps you can convince yourself that this is allowed here?)

Answer 3

$$ \lim _{(x,y) \to (0,2)} (1+xy)^{\frac{2}{x^2+xy}} = \lim_{x \to 0} (1+2x)^{\frac{2}{x^2+2x}} = \lim_{x \to 0} \left((1+2x)^{\frac{1}{x}}\right)^{\frac{2}{x+2}} = e^2 $$