hw2solutions - lim x 16 1 x + 4 = 1 8 . Hence, it follows...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 x - 4 from top and bottom gives us
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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 dont 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 were 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, were 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...
View Full Document

This note was uploaded on 01/18/2012 for the course MATH 2250 taught by Professor Chestkofsky during the Fall '08 term at University of Georgia Athens.

Page1 / 2

hw2solutions - lim x 16 1 x + 4 = 1 8 . Hence, it follows...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online