Looking for help 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

1) you dont have to remove the zero.

remember that lim as x aproaches 0, of [sin(x)] / x = 1

therefor, when x aproaces 0,  lim x / sin (x) = 1

so lim (x-->0) of x / sin (3x) = lim (x-->0) [3x/ sin (3x)] / 3 = lim(x-->0) [3x/ sin (3x)]  / 3 = 1/3.

2) lim (sqrt) = sqrt(lim)

you just replace x with 2 and calculate: 2x-1 = 2*2-1=3.

so the innitial limit = sqrt(3) .

3) also lim as x appr. 3 of (3/(x^2 - 6x +9)) =3 /lim as x appr. 3 (x^2 - 6x +9)) = 3/ 3^2 -6*3+9 = 3/9-18+9 = 3/0

now it depends on HOW x approaches 3

if x<3 then the result is - inf.

if x> 3, then the result is + inf

the basic rule is that 1/ +0 = + inf     and 1/-0 = - inf. :)

4)

if x approaches 0, then 1 / x^2 = + inf, because x^2 is always positive , the result of the limit is + inf, because no matter how much you add to + inf, the result is still + inf.

putting it all togheter doesnt help much, especially since x --> 0.

multiplying something by 0 ussually doesnt help.

good luck, i hope i  made myself clear enough :)