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Practice Quiz 1c solutions

# Practice Quiz 1c solutions - Practice Quiz IC for Calculus...

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Practice Quiz IC for Calculus I, MATH 1501 (I) Let f ( x ) be given by f ( x ) = 1 - x 1 + x for x 6 = 0. (a) Find all x such that | f ( x ) - f (1) | < 1. (b) Find a δ ( ) > 0 such that | x - 1 | < δ ( ) ⇒ | f ( x ) - f (1) | < . and justify your answer. ANSWERS For (a), x > 0. for (b), here are two ways: | f (1 + h ) - f (1) | = | h | | 2 + h | First Way Apply | a + b | ≥ | a | - | b | with a = 2, b = h to get | 2 + h | ≤ 2 - | h | . Hence | f (1 + h ) - f (1) | ≤ | h | 2 - | h | Now for | h | < 2, 2 - | h | > 0, so | h | 2 - | h | < ⇐⇒ | h | < 2 - | h | ⇐⇒ | h | < 2 1 + This gives δ ( ) = min 2 , 2 1 + . Second Way Inpose the condition | h | < 1, under which 1 < 2 + h < 3 and hence | 2 + h | > 1. Decreasing the denominator gives | f (1 + h ) - f (1) | ≤ | h | 2 - | h | < | h | and so if | h | < , we have what we want. Therefore, δ ( ) = min { 1 , } .

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(II) Which of the following limits exist? If they don’t exist, explain why not. If they do, compute the value of the limit. (a) lim x 0 1 x 2 2 + x - 2 (b) lim x 1 cos ( x - 1) 2 ( x 2 - 1) 2 (c) lim x →- 2 x 2 + 5 x + 6 ( x + 2) 2 (d) lim x →- 2 ( x 2 + 5 x + 6) 2 ( x + 2) 2 ANSWERS (a): - 1, (b) cos(1 / 4), (c) does not exist, (d) 1. (III): Consider the following function defined for x 6 = 0: f ( x ) = 1 x 1 x - 1 sin( x ) .
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Practice Quiz 1c solutions - Practice Quiz IC for Calculus...

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