Question

http://webwork.math.ttu.edu/wwtmp/equations/e9/8f9bdfdcdc64a249a59bea911352f61.png

Answer 1

Is there any way you can factor the denominator of the second term?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

\[\lim_{x \rightarrow 1} ( \frac {1}{x+1} + \frac {1}{x+1}) = \lim_{x \rightarrow 1} \frac {2}{x+1}\]

Answer 6

Now just plug in 1 for x, and you get that the limit is 2/2 = 1.

Answer 7

how did you get x of the numerator?

Answer 8

for the equation on the left?

Answer 9

Well, you have (x-1) on both the numerator and denominator, so they just cancel out.

Answer 10

oh ya haha thanks