Maths Problem

Sorry to ka-chiao the regulars and mods here, even though I have gone MIA for quite a while and haven't contribute much to the forum.

But I really need help for this question, thanks in advance:

Limit of [cos[(x+3)^2] - cos9]/x as x approaches zero.

I need help on the steps, I know I can just cheat and change it to f(x) form and differentiate it, but that is not what is required.

Hi,

cos9, is there a typo? Thanks!

Cheers,
Wen Shih

Hi,

cos9, is there a typo? Thanks!

Cheers,
Wen Shih

No Sir. It is bugging me quite a bit because next Tuesday's Calculus Test includes limits and derivatives. Many thanks in advance.

hiiii teacher!! hey have u tried the symbol of the multiple version on that sector??

double post..... sui hor?!!

=1/x [cos(x^2+6x)cos9 - sin(x^2+6x)sin9 - cos 9]

=1/x { [ 1 - (x^2+6x)^2/2+...]cos 9 - [(x^2+6x) + ...]sin9 - cos 9}

= 1/x { cos 9 - (x^2+6x)^2/2* cos9 - (x^2+6x)* sin 9 - cos 9 +... }

= 1/x { - (x^2+6x)^2/2* cos9 - (x^2+6x)* sin 9}+..

= - x(x+6)^2/2* cos9 - (x+6) sin 9 +...

limit as x tends to 0 ==> - 6 sin9

=1/x [cos(x^2+6x)cos9 - sin(x^2+6x)sin9 - cos 9]

=1/x { [ 1 - (x^2+6x)^2/2+...]cos 9 - [(x^2+6x) + ...]sin9 - cos 9}

= 1/x { cos 9 - (x^2+6x)^2/2* cos9 - (x^2+6x)* sin 9 - cos 9 +... }

= 1/x { - (x^2+6x)^2/2* cos9 - (x^2+6x)* sin 9}+..

= - x(x+6)^2/2* cos9 - (x+6) sin 9 +...

limit as x tends to 0 ==> - 6 sin9

Tks a lot. I will have to think through for a while though. :D

Hi hazelp,

Good work, using trigo identity and Maclaurin's series!

Cheers,
Wen Shih

1/x [ cos (x^2 + 6x +9) - cos 9]

=1/x{-2sin[(x^2 + 6x +18)/2]*sin(x^2 + 6x )/2}

=1/x{-2sin[(x^2 + 6x )/2+9]*sin(x^2 + 6x)/2}

={-2sin[(x^2 + 6x )/2+9]*sin(x^2 + 6x)/2} *(x+6)/2

x*(x+6)/2

={-2sin[(x^2 + 6x )/2+9]*sin(x^2 + 6x)/2} *(x+6)/2

( x^2 + 6x)/2

=-2sin[(x^2 + 6x )/2+9]*(x+6)/2

=-sin[(x^2 + 6x )/2+9]*(x+6)

limit as x tends to 0 ==> - 6 sin9