Question

why does the limit of (x+1)^2 -2 / x as x approaches 0 not exist? (x cannot equal 0)

Answer 1

why does \[\lim \frac{ (x+1)^2 -2}{ x }\] as x ->0 not exist?

Answer 2

(x^2+2x+1-2)/x
(x^2+2x-1)/x
x+2-1/x
x => 0+ => 0+2-inff =>-inf
x => 0- => 0+2+inf =>+inf
so limit does not exist

Answer 3

how did u get x+2-1 (2nd to 3rd step)?

Answer 4

Plug in x=0 and you'll see why (you'll get -1/0) There's no need for algebraic manipulation... if a limit approaches pos or neg infinity, then the limit does not exist.

Answer 5

@sara17 it doesn't matter that the left and right hand limits aren't equal in this case. Neither the left or right limits exist since they both approach infinity.

Answer 6

the question is : Let f(x) de defined by f(x)=\[\lim \frac{ (x+1)^2 -2 }{ x }\] for \[x \neq 0\]. Explain why the limit of this function as x approaches o does not exist

Answer 7

Well then, that's a diff question...

Answer 8

yup, when u answered the question, I remember to include that detail. Really sorry :S

Answer 9

btw, "how did u get x+2-1 (2nd to 3rd step)?"
from this:

(x^2+2x-1)/x
x+2-1/x
comes from this:
\[\Large \frac{ x^2 +2x -1 }{ x } = \frac{ x^2 }{ x } + \frac{2x }{ x } - \frac{1 }{ x }\]

Answer 10

if you approach from the left you get positive infinity, and vice versa

Answer 11

are you talking about extended real numbers?

Answer 12

no, but I understand what agent0smith and sara17 were saying now. Thank you all!

Answer 13

actually, maybe it is all real numbers. Not too sure