Limits (again!)?

Question

The last few posts were really helpful, thanks everyone! BUT two questions still remain:

1) limit as x approaches 0 of: 5x^2Cot^2(4x) (the answer is NOT 0)


2) Limit as x approaches 0 of: 8x^2Sin(9/x) (the answer is NOT 0)


Thanks!

Answer 1

I don't understand.

Why not?

When the x value approaches 0, they multiply to 0.

I don't find anyway the answer is not 0.

5(0)^2 cot^2(0) = 0 (0/1) = 0

8(0)^2sin(infinity) = 0 sin(infinity) = 0

To Sahsjing: cot^2(0) is actually undefine, but how did you get 5/16?

Answer 2

1) limit as x approaches 0 of: 5x^2Cot^2(4x) (the answer is NOT 0)

= 5/16, since cot^2(4x) ~ (4x)^2 as x -> 0.

2) Limit as x approaches 0 of: 8x^2Sin(9/x) (the answer is NOT 0)

= 0 (see my previous answer to the same question you asked.)

Answer 3

1) limit as x approaches 0 of: 5x² (cot²4x)

Lim x→0 {5x²(cot²4x)}

= Lim x→0 {5x²/(tan²4x)}

This is of the indeterminant form 0/0 so L'Hospital's Rule applies. It states that the limit of the quotient is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

= Lim x→0 {10x/[8(tan 4x)(sec²4x)]}

= Lim x→0 {10/[32{(sec²4x) + 2(sec²4x)(tan 4x)]}

= 10/32 = 5/16
____________

2) Limit as x approaches 0 of: 8x²sin(9/x)

= Lim x→0 {8x²sin(9/x)}
Let h = 1/x

= Lim h→∞ {8(sin 9h)/h²}

≤ Lim h→∞ { 8(|sin 9h|)/h² }

≤ Lim h→∞ { 8/h² } = 0

The answer is zero.

Answer 4

The first one is
5/16 as stated... It involves some manipulation, some limit laws and a familiar limit

lim 5x^2 cot^2(4x)
x->0

Rewrite in terms of sin and cos
lim 5x^2 cos^2(4x)/sin^2(4x)
x->0

Using limit laws to break into two limits lim(xy) = lim x lim y
lim cos^2(4x) lim 5x^2/sin^2(4x)
x->0

Can evaluate the cos^2(4x) right away it's 1
lim 5x^2/sin^2(4x)
x->0

Bring out 5 write x^2/sin^2(4x) as [x/sin(4x)]^2
5 lim [x/sin(4x)]^2
x->0

Using limit laws you can take the limit of what's inside the square and just square that. lim f(g(x)) = f(lim g(x))
5 [lim x/sin(4x)]^2
x->0

Multiply top and bottom of inside by 4/4
5 [lim 4x/4sin(4x)]^2
x->0

Square the bottom 4 and bring it out.
5/16 [lim 4x/sin(4x)]^2
x->0

as x->0 4x->0 and we basically have x/sin(x) as our limit which we know to be 1. square that and it's still 1

5/16

For the second one it is 0... you can't use direct substitution... but you can use squeeze.
8x^2 sin(9/x)

Start with the range of sin(9/x)
-1 <= sin(9/x) <= 1

Multiply by 8x^2... since it's x^2 you don't need to worry about inequalities changing

-8x^2 <= 8x^2 sin(9/x) <= 8x^2

Take limits.. you can only do the outer two
lim -8x^2 <= lim x^2 sin(9/x) <= lim 8x^2
x->0

0 <= lim x^2 sin(9/x) <= 0
x->0

Therefore by squeeze theorem
lim x^2 sin(9/x) = 0
x->0

Answer 5

Rewrite limit as 5 [4x / 4sin(4x)]^2 cos^2 (4x).

It is well known that n / sin(n) = 1 as n approaches 0. Therefore:

L = 5 * (1/4)^2 * 1 = 5/16.


For the second one, let w = 1/x.

lim 8 sin(9w) / w^2
w->inf

Since sin(9w) is bounded between +-1, L=0. Moreover,

lim sin(9w) / w^2
w-> -inf.

is equal to

lim sin(9w) / w^2
w-> +inf.

so the limit exists.

Answer 6

I think the answer is 5/16. ! may be too naive.

the cos^2(4x) is no pioblem -> 1

5x^2/sin^2(4x) is 0/0 could use L'Hopital, but hard.
expand sin(4x)=4x + (4x)^3 +...
square that 16x^2 + higher order terms

I'll be interested in this one