Compute Limit Using Limit Properties

Let {a_{n}} be a convergent sequence with limit L ∈ ℝ. Use properties of limits to compute $\displaystyle \lim_{n \rightarrow \infty } \frac{ {a}^2_{n} -1 }{{a}^2_{n} +1}} $

Attempt:

1. "Distribute" limit to numerator and denominator (allowed since {a_{n}} is convergent, so {a_{n}}^2-1 and {a_{n}}^2+1 are convergent.

2. "Distribute" limit to {a_{n}}^2 and -1 in the numerator and likewise for {a_{n}}^2 and 1 in the denominator.

3. Since {a_{n}} converges to L, we know {a_{n}}^2 converges even faster to L. Also, we know lim 1 is 1 and lim -1 is -1.

4. Thus, we have (L-1)/(L+1)

Question:

Is this an acceptable solution? I'm not sure if it's okay to have the answer in terms of "L".

Any help would be appreciated.

Re: Compute Limit Using Limit Properties

Hey divinelogos.

Shouldn't a_n^2 converge to L^2 if a_n converges to L?

Re: Compute Limit Using Limit Properties

That makes sense, yes.

So we would have

((L^2)-1)/((L^2)+1)

= (L+1)(L-1)/(L+1)

= L-1

Is this the correct approach in general though? That is, from step 1 where I "distribute" the limit?

Re: Compute Limit Using Limit Properties

Because it is convergent and because its not of an indeterminate form, then this should work.

The distribution should work since you don't get 0/0 (i.e. indeterminant form).

Re: Compute Limit Using Limit Properties

Ok, is there a way to get a "cleaner" limit though? I'm not sure if you can have the answer depend on L.

Also, how do we know we don't get something messed up in the denominator with the distribution or an interderminate form? For example, what if L=-1? Then the denominator would equal 0. Is this okay since a^2 converges to L^2?

Also, to take care of the indeterminate forms, is it okay because the only two indeterminate forms are +-infinity/+-infinite and 0/0, and we know 0's won't pop up because of the + and - 1's, and infinities won't because we know the sequence converges and not diverges?

Re: Compute Limit Using Limit Properties

What do you mean by cleaner? The [L^2 - 1]/[L^2+1] is pretty clean (Also so you have to make sure L^2 != -1 if you have complex numbers).