275_exam2_solution

# X 5 b lim x 2 3 x 5 lim x 2 x 3 5 x since lim x 1 x 0

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- x = 5 . (b) lim x →∞ 2 3 x + 5 = lim x →∞ 2 /x 3 + 5 /x Since lim x →∞ 1 /x = 0 we have lim x →∞ 2 /x = 0 and lim x →∞ 3 + 5 /x = 3 (using the sum and product rule of limits). Then the limit of the ratio is the ratio of the limits (since the limit of the denominator is not zero), so lim x →∞ 2 3 x +5 = 0. Another way to prove this would be by the application of the squeezing principle. If x > 0 then 0 < 2 3 x + 5 < 2 3 x . Since 1 /x 0 as x → ∞ , this will also be true for 2 3 1 x (by the product rule of limits). Another proof would be by using the ε - δ definition (with a large constant c in place of δ , since we take the limit at ). 2

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5. Suppose that the function f ( x ) is continuous on the open interval (0 , 1), and we also have lim x 0 + f ( x ) = 0, lim x 1 - f ( x ) = 1. (a) (Show that there is a continuous function g ( x ) on [0 , 1] for which g ( x ) = f ( x ) for every x (0 , 1). Define the function g ( x ) on [0 , 1] as follows: g ( x ) = 0 if x = 0 f ( x ) if 0 < x < 1 1 if x = 1 This is continuous for every x (0 , 1) (since it’s equal to f which is continuous by assumption). It is also right-continuous at x = 0 (since lim x 0 + f ( x ) = 0 = g (0)) and the same is true for x = 1. But this means that g is continuous on [0 , 1]. (b) Show that f is bounded in (0 , 1). The function g is continuous on [0 , 1] so by the Extreme Value Theorem it is bounded there: | g ( x ) | < M for some M for x [0 , 1]. But then | f ( x ) | < M is true for all x (0 , 1) since g ( x ) = f ( x ) there.
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