-1

How I can prove that the limit is equal to 0?

$$\lim \frac {\sin(y^3x)}{x^2+y^2} =0 $$

when $x\rightarrow 0$ and $y\rightarrow 0$,

Its easy to see that the limit is 0, but how I can prove it?

Thank you.

3SAT
  • 7,627
BAM
  • 433
  • 1
    Hint: Recall that $\forall t\in \mathbb R\left(|\sin(t)|\leq |t|\right)$. – Git Gud May 26 '16 at 10:08
  • Mathjax wise, you might want to mimic some ideas from this: $\lim \limits_{(x,y)\to (a,b)}\left(\dfrac{\sin(f(x,y))}{g(x,y)}\right)$. – Git Gud May 26 '16 at 10:11
  • "Its easy to see that the limit is 0, but how I can prove it?" Sorry but how do you see this, easily or not? Explaining this could have the collateral advantage of adding some context to your question. – Did May 26 '16 at 10:18

1 Answers1

0

You can use directly the definition of limit: for every $\varepsilon>0$ we must have a $\delta>0$ such that for every $x,y$ with $0<|x-y|<\delta$ one has $|\frac{\sin(y^3 x)}{x^2+y^2}|<\varepsilon$.

Well, observe that since $|\sin(z)|\leq |z|$ for every $z\in \mathbb{R}$ we have $$|\frac{\sin(y^3 x)}{x^2+y^2}|\leq \frac{|y^3 x|}{x^2+y^2}.$$

On the other hand, it is always true that $|x|\leq \sqrt{x^2+y^2}$, the side of a square triangle is always less than the hypothenuse and similarly we have $|y|\leq \sqrt{x^2+y^2}$. Hence,

$$|\frac{\sin(y^3 x)}{x^2+y^2}|\leq |\frac{(x^2+y^2)^2}{x^2+y^2}| = x^2+y^2.$$

The latter is the distance square of $(x,y)$ to the origin, i.e. the point we wish to compute the limit to.

Recall that $\varepsilon >0$ is fixed and we need to find $\delta$ with $0<\|(x,y)\|<\delta$ such that the above is less than $\varepsilon$ (here, $\|\cdot\|$ denotes norm). Hence, it suffices to take $\delta =\varepsilon^{1/2}$. Since $\varepsilon>0$ is arbitrary, the continuity follows (by application of the definition of continuity).

Another way to compute this limit is using a polar change of variables. By finding a lower and upperbound which both converge to 0. That is $$\frac{-|y^3 x|}{x^2+y^2}\leq \frac{\sin(y^3 x)}{x^2+y^2} \leq \frac{|y^3 x|}{x^2+y^2}.$$

Now, using $x=r\cos(\theta)$, $y=r\sin(\theta)$, $r>0$ and $\theta \in (0,2\pi)$. $$\lim_{(x,y)\to(0,0)}\frac{|y^3 x|}{x^2+y^2} = \lim_{r\to 0} \frac{r^4 |\sin^3(\theta) \cos(\theta)|}{r^2} = \lim_{r\to 0} r^2 \sin^3(\theta) \cos(\theta) = 0$$ independently of $\theta$ and similarly $$\lim_{(x,y)\to(0,0)}\frac{-|y^3 x|}{x^2+y^2} = \lim_{r\to 0} \frac{-r^4 |\sin^3(\theta) \cos(\theta)|}{r^2} = \lim_{r\to 0} -r^2 \sin^3(\theta) \cos(\theta) = 0.$$ Therefore, $$\frac{\sin(y^3 x)}{x^2+y^2} \longrightarrow 0,$$ as $(x,y)\to (0,0)$.

Martingalo
  • 1,987
  • I didn't read the whole answer, but the last bit, at the least, is incorrect. See this. – Git Gud May 26 '16 at 10:40
  • Sorry this is what I meant, first finding an upperestimate which goes to 0. But it is correct since there is no indeterminacy in the denominator as in the example you give. Here the "polar" limit is always 0 for any $\theta$. – Martingalo May 26 '16 at 11:08
  • In the example it $0$ for every $\theta$ which makes sense. You're still taking a very specific subpath to justify the existence of the limit for all paths. – Git Gud May 26 '16 at 11:13
  • Yes, it's something bounded times limit 0 so polar coordinates work fine here. – Martingalo May 26 '16 at 11:15