Question

PPl Help Me .

Show if this limit exists or not ..

lim ( 1/ x-1) - (2/X to the power 2 -1) as x approaches to 1

..

Answer 1

>_ >

Answer 2

First simplify:

\[\Large \frac{1}{x-1} - \frac{2}{x^2-1}\]

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

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

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

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

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

\[\Large \frac{1}{x+1}\]

Answer 3

So

\[\Large \frac{1}{x-1} - \frac{2}{x^2-1}\]

simplifies to

\[\Large \frac{1}{x+1}\]

Answer 4

To evaluate the limit of this new simplified expression, just plug in x = 1 and simplify

Answer 5

:l Thanx sir TT^TT .. so we just apply the conjugate to get this question done ..

never thought about it =.='

thanx X]

Answer 6

that's one way to look at it, but I'm really using the difference of squares to factor

then I'm combining the fractions (by making the denominators the same) and simplifying as much as possible

Answer 7

yes Sir 8] .. thanx again X3

Answer 8

im a fan of urs noW ^_^

Answer 9

yw

Answer 10

^_^