Limit

[Haftred]

Find the limit algebreically:

lim ((x-1)/(x^2-1))
x --> 1

Thanks

[courtrigrad]

Hello.

To find your limit we cannot just set x equal to 1. This would make the expression indeterminate, i.e. 0/0. Hence we have to use L ' Hopital's Rule.

lim f ' (x) / g ' (x) = 1 / 2x. Setting x = 1 we get 1/2 for the limit.
x-->1

If you want to solve it algebraically:

lim (x-1)/((x^2 - 1) = lim (x-1)/(x-1)(x+1) = lim 1 / (x+1) = 1/2
x-->1 x-->1 x-->1

[Hurkyl]

That's not doing it algebraically. The questions wants you to do something algebraic with (x-1)/(x^2-1)

[Corneo]

Algebra is the real key here. Factor the denominator

x^2-1=(x-1)(x+1)

Then we have

\lim\limits_{x \to 1} \frac {x-1}{x^2-1} = \lim\limits_{x \to 1} \frac {x-1}{(x-1)(x+1)} = \lim\limits_{x \to 1} \frac {1}{x+1} = \frac {1}{2}