Calculating double limit $\lim_{(x,y) \rightarrow (0,0)} (1+xy)^{\frac{1}{x}}$

Question

Tried to evaluate $\lim_{(x,y) \rightarrow (0,0)} (1+xy)^{\frac{1}{x}}$. I know it suppose to be 1. Used a few different methods (sandwich, playing with the terms, using the formal definition and so on), without success.

Would appreciate any help, an hint would be better than full solution.

(Note: part of a basic calculus course)

Thanks

Answer 1

Since $ z/(1+z) \leqslant \log(1 +z) \leqslant z$ for $z > -1$ we have for $(x,y)$ in a neighborhood of $(0,0)$,

$$\frac{y}{1+xy} \leqslant \frac{\log(1+xy)}{x} \leqslant y,$$

and by the squeeze theorem

$$\lim_{(x,y) \to (0,0)}\frac{\log(1+xy)}{x} = 0 \\ \implies \lim_{(x,y) \to (0,0)}(1+xy)^{1/x} = \lim_{(x,y) \to (0,0)}\exp \left(\frac{\log(1+xy)}{x}\right)= 1$$

Answer 2

Let $y=ax, a\in\Bbb{R}$ \begin{align} \lim_{x\to0,y\to0}(1+xy)^{1/x}&=\lim_{x\to0}(1+ax^2)^{1/x}\\ &=\lim_{u\to\infty}[(1+\frac a{u^2})^{u^2}]^{1/u}\tag{Let $u=\frac 1x$}\\ &=\lim_{u\to\infty}e^{a/u}\\ &=1 \end{align}