# Is this correct? Does no limit exist?

I've recently been presented with the following problem:

(b) (3 marks) Now consider the function $g: \mathbb{R}^2 \rightarrow \mathbb{R}$ where

$$g(x, y) = \begin{cases} \frac{\sin(2x^2+2y^2)}{x^2+y^2},& (x, y) \neq (0,0) \\ a,& (x, y) = (0,0) \end{cases}$$

For what value(s) of $a$, if any, is $g(x, y)$ continuous at $(0, 0)$?

And I believe there is no values of a which satisfy continuity. I've taken two limits which are analogous for the Y variable, which describe 4 approaches to the point in question:

$$\lim_{x,0\to0,0} \frac{\sin(2x^2+2(0)^2)}{x^2+(0)^2} = \lim_{x\to0} \frac{\sin(2x^2)}{x^2}$$

I'll skip the evidence we can use L'Hospitals here, but they both converge to 0 (numerator and denominator), therefore applying the rule for this single variable limit:

$$\lim_{x\to0} \frac{4x\cdot\cos(2x^2)}{2x} = \lim_{x\to0} 2\cdot\cos(2x^2) = 2$$

So on this particular approach, $a = 2$ would make the function continuous. However, note that when you take the approach $x = y$, you yield the following (Utilizing product of limit laws):

$$\lim_{x,x\to0,0} \frac{\sin(2x^2+2(x)^2)}{x^2+(x)^2} = \lim_{x\to0} \frac{\sin(4x^2)}{2x^2} = \frac{1}{2}\cdot \lim_{x\to0}\frac{\sin(4x^2)}{x^2}$$

Again we apply L'Hospitals Rule:

$$\frac{1}{2}\cdot\lim_{x\to0} \frac{8x\cdot\cos(4x^2)}{4x} = \frac{1}{2}\cdot\lim_{x\to0}2\cdot\cos(4x^2) = \lim_{x\to0} \cos(4x^2) = 1$$

From this we find a separate value that would also make the function continuous at the point 0,0, so there is no limit that exists. Is this right? According to online calculators there is only one limit, 2, but this path wherein x = y seems to hold up being different...

Can someone poke a hole in my work for me please so I can realise my error?

• As for why the limit is truly $2$, which as far as I can tell isn't what you're asking, one need only note that $(x, y)\to (0,0)$ implies $x^2 + y^2 \to 0$, and then do the substitution $x^2 + y^2 \mapsto t$ to get $g(x, y) = \frac{\sin(2t)}{t}\to 2$. This is basically polar coordinate conversion, but without throwing around a lot of fancy words. Commented Oct 2, 2017 at 11:19
• ...L'Hopital being of course quite inappropriate in this context, being even a logical fallacy.
– Did
Commented Oct 2, 2017 at 11:22
• I’d like to commend you on composing this wonderful example of how to ask a homework-prompted question! Commented Oct 2, 2017 at 12:39
• You made a mistake in your last equation, the denominator should be $2x$ and not $4x$. Then you will obtain the correct limit. Commented Oct 6, 2017 at 22:00

Again we apply L'Hospitals Rule:

$$\frac{1}{2}\cdot\lim_{x\to0} \frac{8x\cdot\cos(4x^2)}{\color{red}2x} = \frac{1}{2}\cdot\lim_{x\to0}\color{red}4\cdot\cos(4x^2) =\color{red}2 \lim_{x\to0} \cos(4x^2) = \color{red}2$$

• Oh, my very large mistake. Thank you so much!
– user431606
Commented Oct 2, 2017 at 11:18
• Now, if I may ask... how else should I go about proving the limit is therefore 2? Conversion to polar co-ordinates (x = rcos(theta), y = r sin(theta))?
– user431606
Commented Oct 2, 2017 at 11:19
• polar coordinate should work as commented by Arthur. Commented Oct 2, 2017 at 11:21
• @PixelRain Because of all the $x^2 + y^2$ in the definition of $g$, that sounds like a very good idea. Commented Oct 2, 2017 at 11:21
• Yeah, thank you all so much, just did the conversion and got alot of nice cancellation, one round of L'Hospitals later and got the limit in concrete. Thank you all so much!
– user431606
Commented Oct 2, 2017 at 11:22

I think the best approach is to just make the substitution $u = x^2 + y^2$.

Then $g$ is continuous at $(0,0)$ if and only if $$\lim_{u\to0^{+}}\frac{\sin(2u)}{u}=a$$ but then, since $$\lim_{u\to0^{+}}\frac{\sin(2u)}{u}=2\left(\lim_{u\to0^{+}}\frac{\sin(2u)}{2u}\right)=2(1)=2$$ we get continuity at $(0,0)$ if and only if $a = 2$.

• Wouldn't it evaluate to 2, given $\sin(2u)$ when derived would return $2cos(2u)$ and the denominator would become 1 after a derivative has been taken, with the limit of cos approach 1 making it 2?
– user431606
Commented Oct 2, 2017 at 11:26
• Yes, thanks -- I was too quick with the mental calculation. Fixed it. Commented Oct 2, 2017 at 11:27
• The question was "Is this right?" and "Can someone poke a hole in my work for me please so I can realise my error?" While this technically answers the question by giving the right answer, this doesn't seem like a good answer to the question.
– JiK
Commented Oct 2, 2017 at 13:11
• The hole was already identified in another answer. More important for the OP is to see a better way (so that in future problems, the OP can benefit from an awareness of the option shown in my answer). Commented Oct 2, 2017 at 13:34
• I wish I could combine the two answers honestly, as they both compliment eachother to actually answer the problem in its entirety. I chose this one afterwards based on the fact others may wish to see how to answer it in future.
– user431606
Commented Oct 3, 2017 at 12:49