4 Questions on finding LIMITS of functions?

Question

Please explain how you got to the answer.

Any help would be appreciated!

Find the limit:

1) lim as x approaches 0 of (x / sin(3x))

How do I remove the zero from the denominator?

2) lim as x appr. 2 from the right of (sq.rt. (2x-1))

3) lim as x appr. 3 of (3/(x^2 - 6x +9))

Again, how do I get rid of the zeros in the denom? The bottom polynomial factors out to (x-3)^2, but how do I go further?

4) lim as x appr. 0 of (3x + 2+ (1/x^2))

I put it together into one fraction ((3x^3 + 2x^2 + 1)/x^2) but I don't know if that will help at all.

Thank you for any help you can offer!

Answer 1

trhe whole secret of finding limits is to avoid letting the denominator get to zero if you possibly can.

Let's look at (1).

By Taylor's theorem sin 3x = 3x - (3x)^3/3! + higher powers of x

now divide your function above and below by x

so you get 1/(3 - 9 x/3! + etc )

and now let x -> 0 so your function -> 1/3

(2) no problem

(3) denom does go to zero, so no limit

(4) same here, you can't get rid of x->0 denominator