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hw2solutions

# hw2solutions - lim x → 16 1 √ x 4 = 1 8 Hence it...

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Math 2250 Written HW #2 Solutions 1. Find the following limit, or explain why the limit doesn’t exist: lim h 0 6 h + 1 - 1 h . Answer: Notice that both the numerator and the denominator go to zero as h 0, so we can’t just use the limit laws to evaluate the limit. In this case, it turns out that the right thing to do is to rationalize the numerator: lim h 0 6 h + 1 - 1 h = lim h 0 6 h + 1 - 1 h · 6 h + 1 + 1 6 h + 1 + 1 = lim h 0 (6 h + 1) - 1 h ( 6 h + 1 + 1) = lim h 0 6 h h ( 6 h + 1 + 1) . Canceling h from both numerator and denominator yields lim h 0 6 6 h + 1 + 1 = 6 2 = 3 , so we can conclude that lim h 0 6 h + 1 - 1 h = 3 . 2. Find the following limit, or explain why the limit doesn’t exist: lim x 16 x - 4 x - 16 . Answer: Again, the numerator and denominator both go to zero as x 16, so we need to get a bit clever. In this case, notice that we can factor the denominator as x - 16 = ( x + 4)( x - 4) . Therefore, lim x 16 x - 4 x - 16 = lim x 16 x - 4 ( x + 4)( x - 4) . Canceling

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Unformatted text preview: lim x → 16 1 √ x + 4 = 1 8 . Hence, it follows that lim x → 16 √ x-4 x-16 = 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 →-2-3( 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 →-2-3( x + 3) | x + 2 | x + 2 = lim x →-2-3( x + 3)-( x + 2) x + 2 = 3 · 1 · (-1) =-3 . 2...
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