Or x eventually the main term of the polynomial will

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→ ∞ or x → -∞ . (Eventually the main term of the polynomial will dominate, so you will need to find the limit of cx n as x → ±∞ .) For rational functions the limits at infinity can also be a real number, this will happen if the degree for the numerator is at most as big as the degree of the denominator. Example 3. Show that lim x →∞ x 2 - 3 x +4 2 x 2 +4 x - 5 = 1 2 . Solution. In order to find the limit at , it is enough to evaluate the function for large values of x . If x > 0 then x 2 - 3 x +4 2 x 2 +4 x - 5 = 1 - 3 x + 4 x 2 2+ 4 x - 5 x 2 whenever both sides make sense. But now the numerator has a limit equal to 1 as x → ∞ and the limit of the denominator as x → ∞ is 2 so the limit of the ration is 1 2 . One can also show that if the rational function f ( x ) = p ( x ) q ( x ) is not defined at x = c (because q ( c ) = 0) then lim x c - f ( x ) and lim x c + f ( x ) will always exist (but might be infinite). Example 4. Show that lim x 2 - x 2 +2 x - 2 = -∞ and lim x 2 + x 2 +2 x - 2 = . Solution. lim x 2 x 2 + 2 = 4 and the one-sided limits of 1 x - 2 are from the left and from the right. From this the statement follows. Practice problems 1. Write out the definition of the following statements using inequalities: (a) lim x 2 + g ( x ) = (b) lim x →∞ h ( x ) = -∞ 2. Find the following limits (you may use the limit laws) (a) lim x 2 1 ( x - 2) 2 (b) lim x 4 - x 4 - x (c) lim x →-∞ 1 x 3 3. Assume that lim x 2 - f ( x ) = and lim x 2 - g ( x ) = 5. Show (with ‘ ε - δ ’ proofs) that (a) lim x 2 - f ( x ) 2 = (b) lim x 2 - f ( x ) g ( x ) = (c) lim x 2 - g ( x ) - f ( x ) = 2
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(d) lim x
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