infinity isn't an actual value itself it describes how big something is

so what it's describing is what happens when x get super super big

well if x is on the bottom of a fraction then we can essentially say that the fraction is 0 because x is so small it's pretty much negligible so we say it's 0

Reply 6

RDKGames

Study Forum Helper

Original post by FrostedStarzzz

http://pmt.physicsandmathstutor.com/download/Maths/A-level/C2/Topic-Qs/OCR-Set-1/C2%20Integration%20MS.pdf

Question 2 b

\displaystyle I= \int_{2}^{\infty}x^{-3}.dx = \lim_{a \rightarrow \infty} \int_{2}^a x^{-3} .dx = \lim_{a \rightarrow \infty}\left[ -\frac{1}{2x^2} \right]_2^a

So we get

\displaystyle I = \lim_{a \rightarrow \infty} \left( -\frac{1}{2a^2} \right) +\frac{1}{8}

Now what is

\displaystyle \lim_{a \rightarrow \infty} \left( -\frac{1}{2a^2} \right)

??

In other words, what value does

-\frac{1}{2a^2}

approach as

a

tends to infinity?

Side note: it might be tempting for many students encountering infinity in this context to just sub it in, but infinity is not a number so we cannot just sub it in. We need to examine what happens to our function as our variable TENDS TO infinity, which is what I've shown here.