Question

Please Help:
lim as x approaches 0
((sinx)(1-cosx))/(-3x^6)

Answer 1

=(sin(x)-sin(x)cos(x))/(-3x^6))
=(sin(x)-sin(2x)/2)/(-3x^6). takes the series expansion of sin(x).
=((x-x^3/6+O(4)-(x-2x^3/3+O(4)))/(-3x^6)
=(x^3/2+O(4))/(-3x^6)
=-1/(6x^3)-O(4)/(6x^6)
=-infinity.

Answer 2

what is the series expansion of sinx?

Answer 3

the series expansion about 0 is x-x^3/6+x^5/120-x^7/5040+...

Answer 4

is there a way to do it without series expansion? I know we learned the rule of sinx/x = 1 but when I apply that the other section is still 0 over 0

Answer 5

you can do it by taking two derivatives instead of one

Answer 6

well I would do that but we have to show all work and for limits at zero we don't use derivatives

Answer 7

\(\dfrac{\sin(x)\cdot(1-\cos(x))}{-3x^{6}} = \dfrac{\sin(x)}{x}\cdot\dfrac{1-\cos(x)}{-3x^{5}} = \dfrac{\sin(x)}{x}\cdot\dfrac{1-\cos^{2}(x)}{-3x^{5}\cdot(1+cos(x))}\)

It is enough.

Answer 8

or would I be able to break it up into
lim (sinx)/(x^6) + lim (1-cosx)/-3 and get infinity +0

Answer 9

that works

Answer 10

really? so that is a legitimate rule?

Answer 11

actually, the limits would have to be multiplied. nvm.

Answer 12

what tk said works though

Answer 13

im following tks but I cant see the whole thing

Answer 14

Well, forget that last part. It doesn't get us anywhere. Splitting off the sine simplifies the issue, but doesn't settle the matter.

Answer 15

On Third thought

\(\dfrac{\sin(x)}{x}\cdot\dfrac{1-\cos^{2}(x)}{-3x^{5}(1+\cos(x))} = \dfrac{\sin(x)}{x}\cdot\dfrac{\sin^{2}(x)}{-3x^{5}(1+\cos(x))}\)

\(= \dfrac{\sin(x)}{x}\cdot\dfrac{\sin^{2}(x)}{x^{2}}\cdot\dfrac{1}{-3x^{5}(1+\cos(x))}\)

NOW! We have settled the matter. The two sine pieces are one (1). The last piece BLOWS UP!

Answer 16

Whoops! That last exponent on x in the denominator should be 3. The result is unaltered.

Answer 17

so the answer is 1?

Answer 18

?? How did you conclude that? "BLOWS UP" is usually not good. There is no limit as x approaches zero (0).

If it were either or both of the sine pieces, yes, it would be just one (1).

Answer 19

so that third piece doesn't exsist causing the rest of the limit to not exsist?

Answer 20

it exists; it is equal to negative infinity.

Answer 21

The other two limits are FINITE. They are easily overcome by one piece that is not finite.

Answer 22

okay i get it now! thanks guys so much!

Answer 23

Infinity is not a place. The limit CANNOT be "infinity" in the Real Numbers. "Infinity" is NOT a Real Number. Besides, in order for a limit to exist, both it's left- and right-hand limits must agree. Not the case on this one.

Approach from the positive side and it wanders off without bound in the negative direction.
Approach from the negative side and it wanders off without bound in the positive direction.

The limit does not exist.