Is this proof of limit existence correct?

Question

So I have to prove the existence of the following: $\lim \limits_{(x,y) \to (0,0)} \frac{-e^{xy} + 1}{xy}$

First I attempt to find $\lim \limits_{(x,y) \to (0,0)} \frac{1}{xy}$, so should this not exist I can assert that the limit does not exist.

Looking at the plot of this function I can see that closing on to the origin the path $y = x$ approaches $\infty$ , and the path $y = -x$ approaches $-\infty$, so I think it's safe to assume that this limit does not exist.

To prove it I write the following:

$\lim \limits_{(x,y) \to (x,x)} \frac{1}{xy}=$ $\lim \limits_{(x,y) \to (x,x)} \frac{1}{x^2}=\infty$

$\lim \limits_{(x,y) \to (x,-x)} \frac{1}{xy}=$ $\lim \limits_{(x,y) \to (x,-x)} \frac{1}{-x^2}=-\infty$

Is this correct? Most methods usually replace $(x,y)$ with $(x, 0)$, $(x,mx)$, etc. so I had doubts as to whether this would be a valid answer.

Thank you in advance.

Answer 1

You have $$\lim_{t\to 0}{1 - e^t\over t} = -\exp'(0) = -1.$$ and $\lim_{(x,y)\to(0,0)} xy = 0.$ What does this say to you?

Answer 2

In This Answer, I showed using the limit definition of the exponential function, the fact that $\left(1+\frac xn\right)^n$ monotonically increases, and Bernoulli's Inequality that the exponential function satisfies the inequality

$$e^z\ge 1+z \tag 1$$

Then, letting $z=-t$, we see from $(1)$ that

$$e^{-t}\ge 1-t\implies e^t \le \frac{1}{1-t} \tag 2$$

for $t<1$. Putting $(1)$ and $(2)$ together yields

$$\frac{1}{1-z}\ge e^z\ge 1+z \tag 3$$

Using $(3)$, we have

$$-\frac{1}{1-xy}=\frac{-xy}{xy(1-xy)}\le\frac{1-e^{xy}}{xy}\le\frac{-xy}{xy}=-1$$

Using the squeeze theorem yields the limit

$$\lim_{(x,y)\to (0,0)}\frac{1-e^{xy}}{xy}=-1$$