Multivariable Exponential Limit Priblem

Question

I am trying to find the following limit $$\lim \limits_{(x, y) \to (0, 0)} \frac{(e^x-1)(e^y-1)}{x+y}$$ I have tried approaching along the lines $y = -x + mx^2$ but I'm getting a zero instead of a value which depends on $m$

Answer 1

Let us calculate, with your suggested $y=x^2-x$. So $$\frac{(e^x-1)(e^y-1)}{x+y} = \frac{e^{x^2}-e^x-e^{x^2-x}+1}{x^2}.$$ Now find the limit as $x\to 0$. One can use L'Hospital's Rule, or find the first few terms of the Maclaurin expansion. These give, for the top $$1+x^2-1-x-\frac{x^2}{2}-1-(x^2-x)-\frac{(x^2-x)^2}{2}+1.$$ Simplify. So the limit along this path is $-1$.

Remark: What I actually did to find out what's going on is to rewrite the top as $e^{x+y}-1-e^x-e^y+2$. The ratio $\frac{e^{x+y}-1}{x+y}$ is nicely behaved. So all we are concerned about is the ratio $\frac{2-e^x-e^y}{x+y}$. The top now has a very nice Taylor series, from which it comes out naturally that taking $y = x^2-x$ will give a non-zero limit, and taking $y = x^3-x$ will make the ratio blow up.

Added: An answer by Nikos_ch, now (I hope only temporarily) deleted, makes the above computation kind of silly. It observes that for $x,y\ne 0$ we can rewrite our function as $$\frac{xy}{x+y}\frac{e^x-1}{x}\frac{e^y-1}{y}.$$ Now we can concentrate on $\frac{xy}{x+y}$ and see instantly what happens along the path $y=x^2-x$.

Answer 2

Try to us the inequalities that hold for small $|x|$:

$$1+x\le e^x\le 1+x+x^2.$$

With these (for small $|x|$)

$$\left|\frac{xy}{x+y}\right|\le \left|\frac{(e^x-1)(e^y-1)}{x+y}\right|\le \left|\frac{(x+x^2)(y+y^2)}{x+y}\right|\le \left|\frac{4|xy|}{x+y}\right|.$$

Now, it is enough to see that

$$\lim\limits_{x\to 0,\ y\to 0}\left|\frac{xy}{x+y}\right|$$

does not exist. See, p.ex.: Does $\lim \frac{xy}{x+y}$ exist at (0,0)?

Answer 3

The numerator is nonzero in the open fourth quadrant, while the denominator vanishes on the line $y=-x.$ There is no hope for a limit, even if you exclude the line $y=-x.$