Limits of quotients - Answer check

[mathbum]

Hey guys, just a heads up, it's been sometime since I've done any highschool math, so I'm a bit rusty.

I'm to find the limit of this function:

lim [(1/x-1) + (1/x+1)] / x
x->0

Since num and denom are both 0, I multiplied the fractions of the numerator by the lowest common denominator, which was (x+1)(x-1), giving me:

[(x+1)(x-1)/(x+1)(x-1)] · [(1/x-1) + (1/x+1)] = after simplifying = [(x+1)/(x+1)(x-1)] + [(x-1)/ (x+1)(x-1)] = after simplifying some more = [(x+1) + (x - 1)]/(x+1)(x-1)

Giving me:
[2x/(x+1)(x-1)]/x

Which I turned into:
[2x/(x+1)(x-1)] · 1/x

2[x/(x+1)(x-1)] · 1/x

Canceling out the X on the numerator and the x on the denominator, I got.

2[1/(x+1)(x-1)] = 2(1/x^2 - 1)

Therefore:
lim 2(1/x^2 - 1)
x-->0

=
lim 2 · (lim 1)/(lim x^2 - 1)
x-->0

2 · (1/-1) = 2 · -1 = -2

Therefore, the limit of the original function is -2.

Amirite?

Thanks for taking the time to read this.

 

[lookagain]

lim [1/(x - 1) + 1/(x + 1)]/ x
x->0

Since num and denom are both 0, I multiplied the fractions of the numerator by the lowest common denominator, which [is] (x + 1)(x - 1), giving me:

[(x + 1)(x - 1)/((x + 1)(x - 1))] · [1/(x - 1) + 1/(x + 1)] = after simplifying = (x + 1)/[(x + 1)(x - 1)] + (x - 1)/[(x + 1)(x - 1)] = after simplifying some more = [(x + 1) + (x - 1)]/[(x + 1)(x - 1)]

Giving me:
2x/[(x + 1)(x - 1)]/x

Which I turned into:
2x/[(x + 1)(x - 1)] · 1/x

Canceling out the x on the numerator and the x on the denominator, I got:

2{[1/[(x + 1)(x - 1)]} = 2[1/(x^2 - 1)]

Therefore:
lim 2[1/(x^2 - 1)]
x-->0

=
lim 2 · (lim 1)/lim(x^2 - 1) . . . . . \(\displaystyle Meaning \lim_{x \to \ 0}{2} \cdot [\lim_{x \to \ 0}{1}}]/[{ \lim_{x \to \ 0}{(x^2 - 1)}}]\)
x-->0 x--> 0 x --> 0

2 · (1/-1) = 2 (-1) = -2

Therefore, the limit of the original function is -2.

[Am I right]?

I added parentheses, brackets, and braces to the above due to the order of operations,
and I spaced out the addition/subtraction symbols.

I got the limit of \(\displaystyle -2\) also.