L'Hôpital evaluating limits for multivariable functions?

Question

Let's say I want to evaluate the following limit:

$$\lim_{(x,y) \to (0,0)} \frac{x^2 + y}{y}$$

and I choose the paths $x = 0$ (which gives me the limit equal to $1$) and I decide on the path $y = x$ also. By substituting $y = x$, does that allow me to use L'Hôpital?

$$\lim_{(x,y) \to (0,0)} \frac{x^2 + y}{y} = \lim_{x \to 0} \frac{x^2 + x}{x} = \lim_{x \to 0} \frac{2x+1}{1} = 1$$

Answer 1

Yes, you can use L'Hôpital though it is not necessary as you can just divide both the numerator and denominator by $x$.

Try to let $y=-x^2$ to prove that the limit doesn't exists.

Answer 2

You can use L'Hôpital's rule on the path limit $$ \lim_{\substack{x\to 0 \\ y = x}} \frac{x^2+y}{y} = \lim_{x\to0} \frac{x^2 + x}{x} \stackrel{\text{H}}{=} \lim_{x\to 0} \frac{2x+1}{1} = 1 $$ However, that doesn't tell you enough to conclude that $\lim_{(x,y) \to (0,0)}\frac{x^2+y}{y}=1$. For as Siong Thye Goh suggests, you can find the limit along another path: $$ \lim_{\substack{x\to 0 \\ y = -x^2}} \frac{x^2+y}{y} = \lim_{x\to0} \frac{0}{-x^2} = 0 $$ Since these two path limits disagree, we know $\lim_{(x,y) \to (0,0)}\frac{x^2+y}{y}$ does not exist.