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Solutions to Practice test 2

# Solutions to Practice test 2 - SOLUTIONS TO PRACTICE TEST 2...

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SOLUTIONS TO PRACTICE TEST 2 1. Find the limits: a) ( ( ( ( ( ( 29 ( 29 2 2 2 4 2 2 2 2 2 4 4 2 2 4 16 lim lim lim lim 2 4 32 2 2 2 x x x x x x x x x x x x x x x - + - + + - = = = + + = - - - b) 3 1 lim 3 x x + = -∞ - c) ( 29 ( ( 29 ( 29 ( 29 ( ( 29 1 1 1 1 1 1 1 1 1 lim lim lim lim 1 2 1 1 1 1 x x x x x x x x x x x x x x - + - - - = = = + = - - - + d) ( 29 5 5 3 2 2 2 3 2 lim lim lim 2 1 x x x x x x x x →+∞ →+∞ →+∞ - + - = = - = -∞ e) 2 2 2 4 4 4 4 3 3 3 5 2 5 2 5 5 lim lim lim lim 0 4 4 x x x x x x x x x x x →+∞ →+∞ →+∞ →+∞ + + = = = = + + f) 1 1 limsin / 2 x x π - = 2. Use the graph to find: a) ( 1 lim x f x - →- b) ( 4 lim x f x + 3. Is the function ( 29 3 2 , 1 2, 1 x x f x x x ≤ - = - - continuous at x = - 1? Explain, briefly, referring to the 3 part definition of continuity. Note that ( 29 ( 29 ( 3 2 3 1 1 1 1 1 lim 2 1; lim 1 x x f and x x + - →- →- - = - = - - = - = - so the 2-sided limit exists at – 1 and = f( - 1), so f is continuous at – 1.

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4. Use the Intermediate Value Theorem to show that 4 5 0 x x - - = has a solution on [1, 2] Let ( 4 5 f x x x = - - This function is continuous since it is a polynomial. Also ( ( 4 4 1 5 1 1 3 ; 2 5 2 2 13 f f = - - = = - - = - Since 0 is a value between 3 and – 13, the IVT guarantees a value x = c for which f(c) = 0.
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Solutions to Practice test 2 - SOLUTIONS TO PRACTICE TEST 2...

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