Question

How to find left hand limit & right hand limit of this limit?
\[\huge{\color{blue}{\lim_{x \rightarrow 1}}}\space {\color{red}{\frac{x}{1-x}.}}\]

Answer 1

plug in 1.0000000000001 and .9999999999999999

Answer 2

well when we plug in something less then 1 but greater than 0 the whole thing is positive and if we plug in something over 1 then the whole thing is negative. so the left hand limit is infinity and the rhl is -infinity

Answer 3

because the bottom goes to 0 and the top is not 0 thus some form of infinity. so we can talk about the left hand limit and the right hand limit at 1, but we cant talk about the limit at 1...

Answer 4

that is also my problem that i gt infinite from one side & - infinite from other side
so the limit of x -> 1 doesn't exist
but in my book there is answer given as infinite:/

Answer 5

You could re-write the problem as:\[\frac{x}{1-x} =\frac{x-1+1}{1-x}=\frac{x-1}{1-x}+\frac{1}{1-x}=-1+\frac{1}{1-x}\]to get more insight. ooks like zzrocker has it right.

Answer 6

looks*

Answer 7

but in my book there is \[\infty\]as answer:/
wt should i do?

Answer 8

i think answer must be
"Limit doesn't exist."

Answer 9

Either the back of the book is wrong in this case, or maybe the question is worded poorly. I could see infinity being the answer if it was asking for a one sided limit.

Answer 10

I agree, limit doesnt exist is correct.

Answer 11

yeah i also think that:)
thank u :D

Answer 12

infinity is not a limit, so if something has a limit of infinity it has no limit

Answer 13

may be!

Answer 14

not maybe, for sure. Infinity is not a number.

Answer 15

:) its just bad notation that we use for some reason

Answer 16

ok:)

Answer 17

the book is still wrong because it goes from infinity and -infinity, so even by bad notation standards the book is still wrong:)

Answer 18

I promise you.

Answer 19

thank u :)