Definite integral approaches infinity

[JasonHathaway]

Homework Statement

$180\int_5^\propto \frac{2}{(4+x^{2})^{3/2}} dx$

Homework Equations

Trigonometric Substitutions: (x=2 tan z).

The Attempt at a Solution

I've computed the integral and ended up with $180 [\frac{x}{2(4+x^2)^{1/2}}]$ from 5 to infinity.

I could've easily computed the limit of 5, but couldn't have found the limit approaches infinity, as I'm always ending up with 0, although my textbook and symolab calculator says that it's 1/2.

 

[Ray Vickson]

Homework Statement

$180\int_5^\propto \frac{2}{(4+x^{2})^{3/2}} dx$

Homework Equations

Trigonometric Substitutions: (x=2 tan z).

The Attempt at a Solution

I've computed the integral and ended up with $180 [\frac{x}{2(4+x^2)^{1/2}}]$ from 5 to infinity.

I could've easily computed the limit of 5, but couldn't have found the limit approaches infinity, as I'm always ending up with 0, although my textbook and symolab calculator says that it's 1/2.

Either use l'Hospital's rule or else note that for x > 0 we have
$$\frac{x}{\sqrt{x^2+4}} = \frac{x}{x \sqrt{1 + 4x^{-2}}} = \frac{1}{\sqrt{1+4 x^{-2}}} \to 1$$
as $x \to \infty$.

 

[SteamKing]

$\propto$ means 'is proportional to' and should not be used to indicate infinity $\infty$

 

[infinitylord]

If it is infinity and not saying proportional to. Then you can calculate the limit to infinity pretty easily.

Take $\frac{x}{2\sqrt{4+x^{2}}}$ and then make the numerator $\sqrt{x^{2}}$ and pull out the $\frac{1}{2}$

So now you can call the whole thing $\frac{1}{2}$$\sqrt{\frac{x^{2}}{4+x^{2}}}$

Now apply L'Hopital's rule.

$\frac{1}{2}$$\sqrt{\frac{2x}{2x}}$

this way, the 2x and 2x cancel to give you $\sqrt{1}$ multiplied by $\frac{1}{2}$. Which was your answer.