{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Example 2 show that lim x 1 x 2 0 solution fix ε 0

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Example 2. Show that lim x →∞ 1 x 2 = 0. Solution. Fix ε > 0. We need c so that if x > c then 1 /x 2 < ε . Choosing c = 1 / ε works. Basic limit laws One can show that the basic limit laws hold for limits at ±∞ . So if lim x →∞ f ( x ) = A , lim x →∞ g ( x ) = B where A, B are real numbers (i.e. not infinite!) then f ( x ) + g ( x ) , f ( x ) - g ( x ) , f ( x ) g ( x ) will also have limits at ( A + B, A - B and AB ), and if B 6 = 0 then f ( x ) /g ( x ) will have a limit A/B . (Similar statements hold at -∞ .) You can also use the Squeezing Principle, this is especially useful if you can estimate complicated functions with simpler ones. If you want to use the limit laws for infinite limits then you have to be a bit more careful: remember that and -∞ are not real numbers. However you can still ‘pretend’ that certain operations can be carried out: + = ’, ‘ ∞ × ∞ = ’, ‘ c × ∞ = ’ if c > 0 and ‘ c × ∞ = -∞ ’ if c < 0, ‘ c = 0’, c + = ’ etc. You should actually use these as rules for limits, not as the outcomes of operations on . That means that you will have statements like this: Assume that lim x p f ( x ) = and lim x p g ( x ) = . Then lim x p ( f ( x ) + g ( x )) = 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
lim x p f ( x ) g ( x ) = lim x p 4 f ( x ) = 0. Note that if f and g have limits at p then we cannot say anything about the limit of f ( x ) - g ( x ) and f ( x ) /g ( x ) there. (It can be basically anything or it might not exist.) Polynomials and rational functions It is not hard to show that for any non-constant polynomial p ( x ) the limit of p ( x ) will be or -∞ ad x
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}