Solving Trigonometry Limit Homework Equation

[songoku]

Homework Statement

limit x to phi/2 of:
(2 + cos 2x - sin x) / (x. cos x + x. sin 2x)

Homework Equations

limit and trigonometry

The Attempt at a Solution

Using differentiation, I got zero as the answer. But I am not able to do without differentiation

(2 + cos 2x - sin x) / (x. cos x + x. sin 2x)

= (2 + 1 - 2 sin²x - sin x) / ( x cos x + 2x sin x cos x)

= (-2 sin²x - sin x + 3) / [x cos x (1 + 2 sin x)]

= [(-2 sin x - 3) ( sin x - 1)] / [x cos x (1 + 2 sin x)]

Stuck there...

Thanks

 

[tiny-tim]

hisongoku!

= [(-2 sin x - 3) ( sin x - 1)] / [x cos x (1 + 2 sin x)]

very good so far …

now (sinx - 1)/cosx = … ? ​

 

[HallsofIvy]

I am going to assume you mean "pi/2".

tiny-tim's method, using a well know trig limit, works. Another way:

In the numerator it is the "sin(x)-1" that makes it 0 at pi/2. In the denominator it is cos(x).

The only way to cancel, then, is to write cos(x) as sqrt{1- sin^2(x)}= sqrt{(1- sin(x))(1- cos(x))}. Now, canceling those two gives

(2sin(x)+3)(\sqrt(1+ sin(x)))}/(x(1+ 2sin(x)))

 

[songoku]

hisongoku! very good so far …

now (sinx - 1)/cosx = … ? ​

Errr..I am not very sure what you mean...(sinx - 1)/cosx = tan x - sec x ??

Based on what HallsofIvy said, it is a well-known trig limit, but I am sorry I don't know the property. Tried to goggle for it but can't find anything. Can you help me? :D

I am going to assume you mean "pi/2".

tiny-tim's method, using a well know trig limit, works. Another way:

In the numerator it is the "sin(x)-1" that makes it 0 at pi/2. In the denominator it is cos(x).

The only way to cancel, then, is to write cos(x) as sqrt{1- sin^2(x)}= sqrt{(1- sin(x))(1- cos(x))}. Now, canceling those two gives

(2sin(x)+3)(\sqrt(1+ sin(x)))}/(x(1+ 2sin(x)))

Oops, yes I meant pi not phi.

Ah, that's the idea to eliminate cos x. I tried to figure out how to eliminate cos x but I failed. Thanks for your idea.

 

[tiny-tim]

hi songoku!

Errr..I am not very sure what you mean...(sinx - 1)/cosx = tan x - sec x ??

(sinx - 1)/cosx = (sinx - 1)(sinx + 1)/cosx(sinx + 1) = … ?

(this is basically the same as HallsofIvy's method )

 

[songoku]

hi songoku!


(sinx - 1)/cosx = (sinx - 1)(sinx + 1)/cosx(sinx + 1) = … ?

(this is basically the same as HallsofIvy's method )

Ah I see. So many methods and I couldn't think even one...whew...

Thanks