{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2.4.5-eg-1

# 2.4.5-eg-1 - x 6 = 3(a In part(a we’re asked for the...

This preview shows page 1. Sign up to view the full content.

Example Let f ( x ) = x 2 - x - 6 | x - 3 | . Evaluate: ( a ) lim x 3 - f ( x ) ( b ) lim x 3 + f ( x ) ( c ) lim x 3 f ( x ) Solution: It’s helpful to first remove the the absolute values in f . Recalling that | y | = ( - y, y < 0 , y, y 0 it follows that the denominator in f satisfies | x - 3 | = ( - ( x - 3) , x - 3 < 0 x - 3 , x - 3 0 = ( - ( x - 3) , x < 3 x - 3 , x 3 . When using this information to rewrite f , the value of x = 3 cannot be used since that makes the denominator zero. So f ( x ) = x 2 - x - 6 - ( x - 3) , x < 3 . x 2 - x - 6 x - 3 , x > 3 = ( x + 2)( x - 3) - ( x - 3) = - ( x + 2) , x < 3 ( x + 2)( x - 3) x - 3 = x + 2 , x > 3 The cancellation of ( x - 3) from the numerator and denominator is valid because
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x 6 = 3. (a) In part (a) we’re asked for the limit as x approaches 3 from the left , so lim x → 3-f ( x ) = lim x → 3--( x + 2) =-5 (b) The limit in part (b) is as x approaches 3 from the right , so lim x → 3 + f ( x ) = lim x → 3 + ( x + 2) = 5 (c) From parts (a) and (b), the left-hand and right-hand limits of f are not the same for x approaching 3, so the limit of f ( x ) as x → 3 does not exist. That is, lim x → 3 f ( x ) = DNE ....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online