# Can someone please prove this limit via the squeeze theorem

Can someone please prove that this limit exists using squeeze theorem? $$\lim_{x,y\to 0,0}\frac{5x^2y}{x^2+8y^2}.$$

Another question I've to ask is for $$y = x^2$$ can we not prove that the limit does not exist? If the case is true then the limit becomes: $$\lim_{x\to 0}\frac{5x^4}{x^2+8x^4}.$$. Can't that limit be solved using L'Hospital's rule and get a value that is not 0? (Just for reference when we approach from both the axis, the limit is 0).

I'm sorry if my framing of the question is messy but

TLDR: I saw this question somewhere which shows that a limit exists but when I tried to use different methods of approaching the limit it gave me different answers.

• For $x\ne 0$, $$\frac{5x^4}{x^2+8x^4}=\frac{5}{1+8x^2}$$ which tends to zero with $x$. Sep 18, 2020 at 12:42
• L'Hospital's rule is not the alpha and omega of limits computation, and anyway it is valid for function of a single variable. Sep 18, 2020 at 12:43
• "TLDR" ? If you write that out, I upvote the question. Sep 18, 2020 at 12:44
• @AnginaSeng "$x^2$" is missing in the numerator Sep 18, 2020 at 12:45
• @Peter too late to edit now! Sep 18, 2020 at 12:57

You have for $$(x,y) \neq (0,0)$$

$$0 \le \left\vert \frac{5x^2y}{x^2+8y^2}\right\vert \le \left\vert\frac{5x^2y}{x^2+y^2} \right\vert\le \frac{5}{2} \vert x \vert \left\vert\frac{\vert x y \vert}{x^2+y^2} \right\vert \le \frac{5\vert x \vert }{2}$$

as $$\vert x y \vert \le \frac{x^2+y^2}{2}$$ for all $$(x,y) \in \mathbb R$$.

As $$\lim\limits_{(x,y) \to (0,0)} \vert x \vert = 0$$, you get the desired conclusion by the squeeze theorem.

• Minor error: there is an $8$ in the denominator.
– user65203
Sep 18, 2020 at 12:49
• This is not an error. I used $\frac{1}{x^2 + 8y^2} \le \frac{1}{x^2 + y^2}$ Sep 18, 2020 at 12:53
• Right, I didn't look carefully.
– user65203
Sep 18, 2020 at 12:55
• I also wanted to ask what exactly was wrong with the method I used, ie using y = x^2 Sep 22, 2020 at 7:58
• The limit exists on the curve $y=x^2$. It is zero. This is coherent with the fact that the limit exists at $(0,0)$. Sep 22, 2020 at 8:12

We have

$$\left|\frac{5x^2y}{x^2+8y^2}\right|=\frac{5x^2|y|}{x^2+8y^2}\le \frac{5x^2|y|+5|y|y^2}{x^2+y^2}= 5|y|\frac{x^2+y^2}{x^2+y^2}=5|y| \to 0$$

or also more simply

$$\left|\frac{5x^2y}{x^2+8y^2}\right| =5|y|\frac{x^2}{x^2+8y^2} \le5|y| \to 0$$

• Yes there are many ways indeed! Thanks
– user
Sep 18, 2020 at 17:35

$$\left|\frac{5x^2y}{x^2+8y^2}\right|=5|y|\left|\frac{x^2}{x^2+8y^2}\right|\le 5|y|\to0.$$

• @mathcounterexamples.net: right. Nor did I specialize the case $x=0$.
– user65203
Sep 18, 2020 at 12:57
• @mathcounterexamples.net: on second reading, I humbly prefer my solution, as it takes less tricks.
– user65203
Sep 18, 2020 at 13:00
• I'm fine if your prefer your solution! Sep 18, 2020 at 13:01
• Why dividing by $x^2$? it seems not necessary to obtain a clear result.
– user
Sep 18, 2020 at 13:09
• @user: you are right, but leaving no trace of $x$ in the numerator makes it more striking. I will update.
– user65203
Sep 18, 2020 at 13:47

0)$$(x,y)\not =(0,0)$$;

1)$$x=0;$$ $$y\not=0;$$ The limit $$=0$$;

2)$$y=0$$; $$x\not=0$$; The limit $$=0$$;

3)$$x,y \not =0$$;

Then

$$|\frac{5x^2y}{x^2+8y^2}| \lt \frac{5x^2|y|}{x^2} =5|y| \rightarrow 0.$$

• Nice solution Peter! Bye
– user
Sep 18, 2020 at 17:35
• user. Thanks, bye. Sep 18, 2020 at 18:14

Hint:

Use polar coordinates: $$\lim_{x,y\to 0,0}\frac{5x^2y}{x^2+8y^2}=\lim_{r\to 0}\frac{r^2\cos^2\theta\cdot r\sin\theta}{r^2\cos^2\theta+8r^2\sin^2\theta},$$ and observe that $$\;|\cos^2\theta\sin\theta|\le 1$$, whereas $$\:\cos^2\theta+8\sin^2\theta=1+7\sin^2\theta\ge 1$$.