Evaluating Limits Algebraically help needed.?

Question

1.) lim (x^3-2x) / (x-2)
x ->2

I know that it is indeterminate and that I have to evaluate algebraically. I just can't figure out what to do with the x^3. I factored it out to x(x^2-2) / (x-2) but I don't know where to go from there. Is that as far as you can take it so it would be DNE?

lim (sq. root 1+x) - (sq. root 1-x) / x
x ->0

I have no problems when there's just one square root, but for some reason when there's two I can't seem to ever get the right answer.

Any help is great. Thanks.

Answer 1

=the numerator has lim=4 and the denominator has limit zero so the limit is infinity.It is not indeterminate
2) multiply by sqrt(1+x)+sqrt(1-x) both terms of the division(it is called the conjugate (a+b)*(a-b) =a^2-b^2))
= (1+x-1+x)(/sqrt(1+x)+sqrt(1-x))*x
simplify x
and you get lim = 2/2=1

Answer 2

1.) This is not indeterminate. As x -> 2, the numerator gets close to 4. As x -> 2- the denominator gets close to 0 through negative values, so the function has left-hand limit
-inf. But as x -> 2+ the denom gets close to 0 through positive values, so the function has right-hand limit +inf.

Since left- and right-hand limits differ, the limit does not exist.

2.) Multiply top and bottom by sqrt(1 + x) + sqrt(1 - x). You get 2/[sqrt(1 + x) - sqrt(1 - x)]. That is continuous at x = 0, so the limit is 2/(1 + 1) = 1.