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(be careful when you are solving this inequality both

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Unformatted text preview: (Be careful when you are solving this inequality: both x- 3 and c are negative.) This means that we can choose δ =- 1 c . 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 • lim x → p f ( x ) g ( x ) =...
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