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

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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
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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
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