8b midterm 2

8b midterm 2 - Practice Midterm 2 Solutions 1. Compute lim...

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Unformatted text preview: Practice Midterm 2 Solutions 1. Compute lim x - 4 3- x 2- 7 x +4 lim x - 4 3- x 2- 7 x + 4 = 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 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 - x 2 = lim x x 2 = 0 So, by the Sandwich Theorem we must also have lim x x 2 sin 1 x = 0 1 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 )....
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8b midterm 2 - Practice Midterm 2 Solutions 1. Compute lim...

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