Hey divinelogos.
Shouldn't a_n^2 converge to L^2 if a_n converges to L?
Let {a_{n}} be a convergent sequence with limit L ∈ ℝ. Use properties of limits to compute
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.
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?