# Two doubts

pbialos
Hi, Today i had a test, and i was wondering if what i did is correct:

I had to tell if the $$:lim_{(x,y) \rightarrow (0,0)} \frac {x*y} {|x|+|y|}$$ exists. What i did was to say $$lim_{(x,y) \rightarrow (0,0)} \frac {x*y} {|x|+|y|}=lim_{(x,y) \rightarrow (0,0)} \frac {x} {|x|+|y|}*y=0$$ because it is bounded multiplied by y that tends to 0.
Is what i did correct???Something tells me it is not because it was too easy.

The second doubt i have is about the convergence of the Integral:
$$\int_{0}^{\infty}\frac {sin(x)} {cos(x)+x^2}$$
My doubt came since the integrand of the series is not always positive. Can i just calculate the convergence of:
$$\int_{0}^{\infty}|\frac {sin(x)} {cos(x)+x^2}|$$ and if it converges, then the original integral also converges???
I would check the convergence of the second integral dividing it between 0-1 and 1-infinity. The integral on the first interval(between 0 and 1) would converge because the integrand is bounded and continuous for x between 0 and 1, and the integral of the second interval(between 1 and infinity) would also converge by comparison with $$\int_{1}^{\infty}\frac {1} {x^2-1}$$.

I would really appreciate any help.
Regards, Paul.

Related Introductory Physics Homework Help News on Phys.org
AKG
Homework Helper
I'm rusty on stuff like convergence, but what you did with the second problem seems right. However, are you sure that $\int _1 ^{\infty }\frac{1}{x^2 - 1}\, dx$ converges?

Last edited:
Galileo
Homework Helper
AKG said:
On the other hand, approach along y = 0, you get:

$$\lim _{(x,0) \to (0,0)} \frac{0x}{|x| + |0|} = \lim _{x \to 0}\frac{0}{|x|} = \infty$$
Since the numerator is zero along that path, the limit will clearly also be zero along that path.

pbialos, I think you have the right idea. It is indeed not difficult, but the phrasing was a bit vague. It's not clear to me what you mean by "because it is bounded multiplied by y that tends to 0."

Try something like:

$$\left|\frac {xy} {|x|+|y|}\right|=|y| \frac{|x|}{|x|+|y|} \leq |y|$$
because $|x|/(|x|+|y|) \leq 1$. Since |y| approaches zero as (x,y) approaches (0,0), the limit is zero.

AKG
Homework Helper
Galileo said:
Since the numerator is zero along that path, the limit will clearly also be zero along that path.
Yeah, I don't know what I was thinking... edited.

pbialos
thank you

Many thanks for your responses. Yes, what Galileo said was my idea, although i expressed my self terribly bad.

Regards, Paul.