How do I find $\lim_{(x,y) \to (0,0)} \frac{\sqrt{xy + 1} -1}{x + y}$?

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Compute $$\lim_{(x,y) \to (0,0)} \frac{\sqrt{xy + 1} -1}{x + y}$$

If you look at the graph of the function, it suggests that the limit does not exist. In fact, the limit DOES NOT exist. But how do I prove that?

When I approach the function from either $x$ or $y$-axis, the limit is equal to $0$. It is also equal to $0$ for every line (except for $x=-y$) passing through the origin.

I tried to approach $0$ on a parabola $y=x^2-x$ but the limit was again $0$.

Any thoughts would be appreciated.

Edit:

I did a mistake when calculating the limit as I approached $0$ on the parabola $y=x^2-x$ the result should be $1$ for that limit proving that the limit does not exist.

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In the OP, it appears that you have analyzed the limit along straight-line and parabolic trajectories.

Let's analyze the case for which the limit is approached along the path $y=-x+x^\alpha$. Then, we have

$$\frac{xy}{x+y}=\frac{-x^2+x^{1+\alpha}}{x^\alpha}=x-\frac{1}{x^{\alpha-2}}\tag1$$

What can you conclude now?


For the record, if the limit is approached along the path $y=-x+x^2$, then with $\alpha=2$ in $(1)$, we find

$$\frac{xy}{x+y}=\frac{-x^2+x^3}{x^2}=x-1\to -1\ne 0$$

from which we can conclude the limit fails to exist.