Limits Question - Basic Calculus?

Question

I should know this but it's been alomst a year since I've done calculus. So how would I go about solving the limits:

1) limit as y approaches 4 of: ((1/y)-(1/4))/(y-4)


2) Limit as x approaches 0 of: 8x^2Sin(9/x)


3) Limit as x approaches 0 of: (1-cos(2x))/5x^2


Thanks!

Answer 1

1) lim y->4 ((1/y)-(1/4))/(y-4)
= lim y->4 ((4-y)/4y)/(y-4) {Taking lcm}
= lim y->4 (-(y-4)/4y)/(y-4)
= lim y->4 -1/(4y)
= - lim y->4 1/(4y)
= - 1/4(4) {Substitute y by 4}
=-1/16

2) lim x->0 8x^2 sin(9/x)
=8 lim x->0 x^2 sin(9/x)
=8 lim x->0 (x^2 sin(9/x))/(1/x) {Dividing by 1/x}
=8 { lim x->0 x^2 } {lim x->0 9[sin(9/x)/(9/x)]}
=8*9{0^2} {1} { since lim of x->0 sinx/x=1}
=0

3) lim x->0 (1-cos(2x))/5x^2
=lim x->0 2 sin^2x/5x^2
=2/5 lim x->0 (sin^2x/x^2)
=2/5 [lim x->0 sinx/x}^2
=2/5(1) { since lim of x->0 sinx/x=1}
=2/5

Answer 2

1) limit as y approaches 4 of: ((1/y)-(1/4))/(y-4)
This is 0/0 so use L'Hospital's rule:
limit -1/y^2/1 = -1/16

2) Limit as x approaches 0 of: 8x^2Sin(9/x) =0
= lim 8x^2/csc(9/x) = 0
3) Limit as x approaches 0 of: (1-cos(2x))/5x^2
= lim 2sin(2x)/10x = lim 4cos(2x)/10 = 2/5

Answer 3

1) limit as y approaches 4 of: ((1/y)-(1/4))/(y-4) = -1/16


2) Limit as x approaches 0 of: 8x^2Sin(9/x) = 0


3) Limit as x approaches 0 of: (1-cos(2x))/5x^2 = 2/5
--------------
Ideas:
1) 1/y - 1/4 = (4-y)/(4y)
2) -8x^2 ≤ 8x^2Sin(9/x) ≤ 8x^2
3) 1-cos(2x) = 2sin^2(x)

Answer 4

1) Multiply by the conjugate (same thing THAT EQUALS ONE but with a switched sign in the middle, e.g.(1/y)+(1/4)/(1/y)+(1/4)), simplify, then substitute 4 for y.

2) Use the exponents rule... 2(8)x^1 Sin(9/x) = 16x Sin (9/x)... uh and I'm guessing undefined?? Probably not, though; sorry.

3) Again, multiply by the conjugate (see 1.)

Hope this helps.

Answer 5

Differentiate top and bottom.

1) (-1 / y^2) / 1

-1 / 16

2) 0 (Not so sure)

3) 2 sin 2x / 10x

4cos2x

Answer: 4

*EDIT* Sahsjing has the correct answer for the third question.

Answer 6

-0.0625
0
0