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Unformatted text preview: lim x → 16 1 √ x + 4 = 1 8 . Hence, it follows that lim x → 16 √ x4 x16 = 1 8 . 3. Determine each of the following limits, or explain why they don’t exist. (a) lim x →2 + 3( x + 3)  x + 2  x + 2 (b) lim x →23( x + 3)  x + 2  x + 2 1 (a) Answer: Since we’re interested in x >2, we know that x + 2 > 0, which means that  x + 2  = x + 2 . Hence, lim x →2 + 3( x + 3)  x + 2  x + 2 = lim x →2 + 3( x + 3) x + 2 x + 2 = 3 · 1 · 1 = 3 . (b) Answer: In this part, we’re interested in x < 2, which means that, for the values of x we care about, x + 2 < 0. Therefore,  x + 2  =( x + 2) , so we have lim x →23( x + 3)  x + 2  x + 2 = lim x →23( x + 3)( x + 2) x + 2 = 3 · 1 · (1) =3 . 2...
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 Fall '08
 CHESTKOFSKY
 Math, Calculus, lim, Elementary arithmetic, Power of two, Sexagesimal

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