Can someone please prove this limit via the squeeze theorem

Question

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 $x$. — Angina Seng, Sep 18 '20 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. — Bernard, Sep 18 '20 at 12:43

mathcounterexamples.net · Accepted Answer · 2020-09-18T13:00:30.393

5

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.

edited Sep 18 '20 at 13:00

answered Sep 18 '20 at 12:43

mathcounterexamples.net

70,018

Minor error: there is an $8$ in the denominator. – Sep 18 '20 at 12:49
This is not an error. I used $\frac{1}{x^2 + 8y^2} \le \frac{1}{x^2 + y^2}$ – mathcounterexamples.net Sep 18 '20 at 12:53
Right, I didn't look carefully. – Sep 18 '20 at 12:55
I also wanted to ask what exactly was wrong with the method I used, ie using y = x^2 – Saurav Chittal Sep 22 '20 at 07:58
The limit exists on the curve $y=x^2$. It is zero. This is coherent with the fact that the limit exists at $(0,0)$. – mathcounterexamples.net Sep 22 '20 at 08:12
So what exactly from my explaination in my question was wrong? Was it the fact that you couldn't apply L'Hopital's rule? Or something else? – Saurav Chittal Sep 22 '20 at 09:27
It is the fact that the limit do exists. – mathcounterexamples.net Sep 22 '20 at 09:35
Yet with my method I'm getting a different limit. I do realize the limit exists as shown by not only by your awesome but several others too. So yes I realize the limit exists. However, I wanted to ask what exactly was wrong with how I calculated the limit at y = x^2. – Saurav Chittal Sep 22 '20 at 10:02

score 2 · Answer 2 · 2020-09-18T13:48:49.053

2

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

edited Sep 18 '20 at 13:48

answered Sep 18 '20 at 12:52

@mathcounterexamples.net: right. Nor did I specialize the case $x=0$. – Sep 18 '20 at 12:57
@mathcounterexamples.net: on second reading, I humbly prefer my solution, as it takes less tricks. – Sep 18 '20 at 13:00
I'm fine if your prefer your solution! – mathcounterexamples.net Sep 18 '20 at 13:01
1

Why dividing by $x^2$? it seems not necessary to obtain a clear result. – user Sep 18 '20 at 13:09
1

@user: you are right, but leaving no trace of $x$ in the numerator makes it more striking. I will update. – Sep 18 '20 at 13:47

user · Answer 3 · 2020-09-18T13:07:06.250

2

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

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

edited Sep 18 '20 at 13:07

answered Sep 18 '20 at 13:01

user

154,566

1

Yes there are many ways indeed! Thanks – user Sep 18 '20 at 17:35

score 2 · Answer 4 · answered Sep 18 '20 at 13:55

2

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

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

answered Sep 18 '20 at 13:55

Peter Szilas

20,344
2
17
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Nice solution Peter! Bye – user Sep 18 '20 at 17:35
user. Thanks, bye. – Peter Szilas Sep 18 '20 at 18:14

Bernard · Answer 5 · 2020-09-18T17:47:30.527

1

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 $\:\cos^2\theta+8\sin^2\theta=1+7\sin^2\theta\ge 1$.

edited Sep 18 '20 at 17:47

answered Sep 18 '20 at 12:47

Bernard

175,478

Can someone please prove this limit via the squeeze theorem

5 Answers5