Solving Limits and Convergence: Doubts and Calculations Explained

[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.

 

[AKG]

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?

 

[Galileo]

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]

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.