Proving a limit equals $0$ using a given theorem

Question

Let $a$ be a real number. Prove that if

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

then we must have $a = 0$.

I would like to solve this using the following fact:

$$\lim_{(x, y) \to (x_0, y_0)} \frac{f(x,y) - \psi(x,y)}{\sqrt{(x - x_0)^2 + (y - y_0)^2}} = 0$$

where $\psi(x, y) = f(x_0, y_0) + f_{x}(x_0, y_0)(x - x_0) + f_y(y - y_0)$.

I'm pretty sure this is the right way to do it, because the exercise just before this one (which I solved) was to show $\lim_{(x, y) \to (0,0)} \frac{ax^2 + bxy + y^2}{\sqrt{x^2 + y^2}} = 0$. This was done by defining $f(x, y) = ax^2 + bxy + y^2$, and just plugging it into the fact above (we end up getting $\psi(x, y) = 0$).

I can't figure this problem out, though. I tried defining $f(x, y) = ax$, and I got $\psi(x, y) = ax$, which didn't get me anywhere. Any help is appreciated.

Answer 1

Note that if $a$ is any nonzero number and the limit exists then $$ \lim\frac{ax}{\sqrt{x^{2}+y^{2}}}=a\lim\frac{x}{\sqrt{x^{2}+y^{2}}}. $$ Next, consider the sequence $x_{n}=1/n$ and $y_{n}=0$. It follows that $$ \frac{x_{n}}{\sqrt{x_{n}^{2}+y_{n}^{2}}}=\frac{1/n}{\sqrt{(1/n)^{2}}}=1 $$ and hence $$ \lim\frac{x}{\sqrt{x^{2}+y^{2}}}\neq0. $$ Therefore, $a$ must be zero.

Answer 2

Let $f(x,y):= \frac{ax}{\sqrt{x^2 + y^2}}$ . For $x \ne 0$ we have

$f(x,x)=\frac{ax}{\sqrt{2}|x|}.$

Hence $ \lim_{x \to 0+}f(x,x)=\frac{a}{\sqrt{2}}$ and $ \lim_{x \to 0-}f(x,x)=-\frac{a}{\sqrt{2}}$.

Conclusion: if $\lim_{(x, y) \to (0,0)} \frac{ax}{\sqrt{x^2 + y^2}} = 0$, then $\frac{a}{\sqrt{2}}=-\frac{a}{\sqrt{2}}$ and threrfore $a=0.$