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8b midterm 2 - Practice Midterm 2 Solutions 3 x2-7 x 4 1...

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Practice Midterm 2 Solutions 1. Compute lim x →- 4 3 - x 2 - 7 x +4 lim x →- 4 3 - x 2 - 7 x + 4 = 0 0 if we plug in 4, which is undefined Simplify lim x →- 4 3 - x 2 - 7 x + 4 = lim x →- 4 (3 - x 2 - 7) ( x + 4) · (3 + x 2 - 7) (3 + x 2 - 7) = lim x →- 4 9 - ( x 2 - 7) ( x + 4)(3 + x 2 - 7) = lim x →- 4 16 - x 2 ( x + 4)(3 + x 2 - 7) = lim x →- 4 (4 + x )(4 - x ) ( x + 4)(3 + x 2 - 7) = lim x →- 4 4 - x 3 + x 2 - 7 = 8 12 = 2 3 2.Compute lim x 0 x 2 sin 1 x . Notice plugging in zero tells us nothing because what is sin( )?? Here we need to be aware of the fact that for any value of x , - 1 sin ( 1 x ) 1. So then we have - x 2 x 2 sin ( 1 x ) x 2 . Now, lim x 0 - x 2 = lim x 0 x 2 = 0 So, by the Sandwich Theorem we must also have lim x 0 x 2 sin 1 x = 0 1
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3. f ( x ) = - 2 x + 1 , x 2 x 2 - 6 x + 5 , x > 2 a). Find lim x 2 + f ( x ) and lim x 2 - f ( x ). lim x 2 + f ( x ) = lim x 2 + x 2 - 6 x + 5 = 2 2 - 6 · 2 + 5 = - 3 And lim x 2 - f ( x ) = lim x 2 - - 2 x + 1 = - 2 · 2 + 1 = - 3 b). Does lim x 2 f ( x ) exist? Why or Why not?
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