Proving Epsilon-Limit for a Sequence with Algebraic Manipulation

[Unassuming]

$$Suppose \lim_n \frac{a_n -1}{a_n +1} = 0. Prove that \lim_n a_n = 1.$$

I am trying to do the algebra so that -a_n < ?? < a_n , but I am having trouble. Am I going about this correctly?

I have also tried to solve each separate side of the inequality. I get a_n < (e+1)/(1-e), but this is not quite fitting.

Can somebody give me a clue please. Thanks

Edit: There is a hint in the book that says to set,

$$\(x_{n}=\frac{a_n-1}{a_n+1}\)$$ and then solve.

I have done this but I don't know what to do next.

 

[HallsofIvy]

I presume the hint said set $x_n= (a_n-1)/(a_n+ 1)$ and solve for x_n. When you did that what did you get? What is the limit of that as x goes to 0?

 

[Unassuming]

I solved for a_n and I then took the lim of both sides so that

lim a_n = lim ( -x_n -1 ) / (x_n -1 ), then it was pretty straightforward.

Assuming I can take the lim of both sides. Can I do that?

 

[HallsofIvy]

Yes, of course. Just use the basic "rules" for limits. If the limits of a_n and b_n exist, and the limit of b_n is not 0, then the limit of a_n/b_n exists.