How to evaluate a multi-variable limit considering $f : \mathbb R^2 \to \mathbb R$?

Question

Let $f : \mathbb R^2 \to \mathbb R$ be defined by $$ f(x, y) = \frac{x^2}{\sqrt{x^2 + y^2}}\,, \quad(x, y) \ne (0, 0). $$ Show that $ \lim_{(x, y)\to (0, 0)} f(x, y) = 0 $ using the $ \varepsilon - \delta $ definition.

Source: MTH 303 - Advanced Calculus/OAU - Harmattan Semester Examinations/2018 - 2019 Academic Session/Multivariable Limit/Q1. (a)

Comment: I am good at evaluating limit as regards $f : \mathbb R \to \mathbb R$ in single-variable calculus, but in multi-variable calculus I am not that good.

I will be glad if an explanation is provided alongside with the calculation. Thanks!

Answer 1

You need to bound $f(x,y)$ appropriately from above.

So, let $\epsilon > 0$:

$$|f(x, y)| = \frac{x^2}{\sqrt{x^2 + y^2}}\leq \frac{x^2+y^2}{\sqrt{x^2 + y^2}} = \sqrt{x^2+y^2} \stackrel{!}{<}\epsilon$$.

So, for $\boxed{\delta(\epsilon) = \epsilon}$ you get for all $(x,y) \neq (0,0)$ with $\sqrt{x^2+y^2} < \delta(\epsilon)$ the desired inequality $|f(x,y)|< \epsilon$, which is exactly the meaning of $f$ having the limit $0$ at $(0,0)$.

Answer 2

How about polar? Get $\dfrac {r^2\cos^2\theta}r=r\cos^2\theta\le r\to0$.

Or you could do it without: $\vert\dfrac {x^2}{\sqrt{x^2+y^2}}\vert\le\vert\dfrac {x^2}{\sqrt {x^2}}\vert=\vert\dfrac {x^2}x\vert=\vert x\vert$, which means we can take $\delta=\epsilon $.

Answer 3

The general idea for such $\epsilon$-$\delta$ proofs is that you want to get an upper bound on $|f(x,y) - 0|$, which you can make $< \epsilon$. One possibility is like I hinted at in the comments: Notice that for any $(x,y) \in \Bbb{R}^2$, \begin{align} |x| = \sqrt{x^2} \leq \sqrt{x^2 + y^2} \end{align} Hence, if $(x,y) \neq 0$, then $ \dfrac{|x|}{\sqrt{x^2+y^2}} \leq 1$. So far none of this is a proof, it is merely "scratch work".

A formal proof goes something like:

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Let $\epsilon > 0$ be arbitrary, and choose $\delta = \epsilon$. Let $(x,y) \in \Bbb{R}^2$ any arbitrary element such that $0 < \lVert (x,y) \rVert < \delta$. Then, \begin{align} |f(x,y) - 0| &= \left| \dfrac{x^2}{\sqrt{x^2 + y^2}} \right| \\ &= \dfrac{|x|}{\sqrt{x^2+y^2}} \cdot |x| \\ &\leq 1 \cdot |x| \\ &\leq \lVert (x,y) \rVert \\ &< \delta \\ &= \epsilon \end{align} (I tried to not skip many steps in the inequalities above, so you should be able to justify each of them)

Thus, we have shown that for any arbitrary $\epsilon > 0$, there exists a $\delta > 0$ (in our case it was simply $\delta = \epsilon$) such that for all $(x,y) \in \Bbb{R}^2$, if $0< \lVert (x,y) \rVert < \delta$ then $|f(x,y) - 0| < \epsilon$. This is precisely what it means to prove \begin{align} \lim \limits_{(x,y) \to (0,0)}f(x,y) = 0. \end{align} This completes the proof.

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Additional Remarks:

Normally, there's no need to be this detailed with the inequalities, but if you're just starting out, you should practice justifying each step carefully, hence I wrote out a detailed proof.

Also, there are often many solutions to the same question, and this is no exception. I think in this particular case, @trancelocation's solution for getting an upper bound on $|f(x,y)|$ is slightly simpler to follow, so you should try to understand that as well.

Answer 4

$$\left|\frac{x^2}{\sqrt{x^2+y^2}}\right|\leq|x|.$$ Now, use the $\epsilon-\delta.$