Question

find the limit(if it exists). (If an answer does not exist, enter DNE.)
lim_{x rightarrow 1+}\frac{|x-1|}{ x-1}

Answer 1

Oooo! It's Pooh! \C:/

Answer 2

\[\lim_{x \rightarrow 1+}\frac{|x-1|}{ x-1}\]

Answer 3

Oh only the top is absolute, ok ok.

Answer 4

You can write |x-1| as a piece-wise, that might help,\[\Large\rm |x-1|=\cases{x-1, &x>1\\ \rm -(x-1), &x<1}\]So when we're approaching from the RIGHT side,
our |x-1| just simply equals x-1, yes?

Answer 5

That make a little sense? :o

Answer 6

\[\Large\rm \lim_{x \to1^+}\frac{|x-1|}{ x-1}=\quad \lim_{x \to1^+}\frac{x-1}{ x-1}\]

Answer 7

Where you at Pooh! Put down the honey!!

Answer 8

lol

Answer 9

XD

Answer 10

lol sorry, yes i understand that part

Answer 11

So then uhhhh.. yah.. before evaluating the limit, it looks like you have a nice cancellation, yes? :o

Answer 12

YES

Answer 13

so do i plug in the 1 for x?

Answer 14

You first make your cancellation,
\[\Large\rm =\lim_{x \to1^+}\frac{\cancel{x-1}}{\cancel{x-1}}\]
\[\Large\rm =\lim_{x\to1^+}1\]And then plug in 1 for any x's in the expression,
Hmm, no x's :3\[\Large\rm =1\]

Answer 15

oh ok

Answer 16

thank you :)