Question

Another Limit question. 1/x+1 - 1/2 On x-1
Lim x -> 1

Answer 1

\[\frac{ 1/x+1 - 1/2 }{ x-1 }\]

Answer 2

(1/x+1) - (1/2)

Answer 3

i am going to guess it is
\[\frac{\frac{1}{x-1}-\frac{1}{2}}{x-1}\]

Answer 4

yes, a little hard to format haha

Answer 5

nope \[\huge \frac{\frac{1}{x+1}-\frac{1}{2}}{x-1}\]

Answer 6

... mb

Answer 7

is it right?

Answer 8

what i wrote i mean

Answer 9

Yes now it is

Answer 10

you want to replace x by 1 but you cannot , because you will get \(\frac{0}{0}\) that means your goal is to cancel the factor of \(x-1\) top and bottom ]
to do this you need to simplify this compound fraction q

Answer 11

it is algebra all the way
first off do the subtraction in the numerator
subtract
\[\frac{1}{x+1}-\frac{1}{2}\]

Answer 12

So you multiply the denominator to match them up?
(2)(X+1)

Answer 13

yes, the denominator will be \(2(x+1)\)

Answer 14

leave it in factored form is fine

Answer 15

\[\frac{ x+1 }{ 2(x+1) } - \frac{ 2 }{ 2(x+1) }\]

Answer 16

yup

Answer 17

x + 1 cancel out and 2 remains

lim x -> 1 = 2

Answer 18

ending up the the \(x-1\) up top that you want to cancel with the bottom

Answer 19

crap yes

Answer 20

whoa hold the phone you still have to subract

Answer 21

you still have a compound fraction as well

Answer 22

\[\frac{ x+1 }{ 2(x+1) } - \frac{ 2 }{ 2(x+1) }=\frac{x-1}{2(x+1)}\] and that is just then numerator

Answer 23

but now it is easy
divide by \(x-1\) and you are left only with
\[\frac{1}{2(x+1)}\] and now you are free to replace \(x\) by \(1\)

Answer 24

Oh okay, I was wondering how you got x - 1 for a sec. But thanks! This really helps :)

Answer 25

yw