find the limit of (1-cos2x-cos4x)/x as x->0 and show all work!

Question

fin the limit of (1-cos2x-cos4x)/x as x->0 and show all work!

Answer 1

We can not apply L' Hopital's rule because it does not satisfy the condition that either it should be

Of the form 0/0 or. Infinite/infinite

 

Limit                 -cos(4x)/x

 

 

x tends to 0

 

It should be -infinite & not zero

Answer 2

We plug in 0 for X and we get cos 2*0= cos 0= 1 and cos 4*0=cos 0=1. Therefore the limit is (1-1-1)/0=-1/0=-∞

The limit does not exist.

Answer 3

If this is in Calculus II, you can use L'Hopital's Rule.  If this is Calculus I, it will take a while.

 

The answer will be 0.  It is really easy if you use L'Hopital's Rule: d/dx for both top and bottom.

(1-cos2x-cos4x) becomes (2sin2x + 4sin4x) and x becomes 1.  As you put 0 for x, it will approach 0.

 

Remember, you can find special trigonometric limits from your textbook (if you are in Calc I):

 

lim     (sin x)/x = 1            lim   (1 - cos x)/x = 0     lim   (1) / (cos x) = 1   

x->0                                x->0                             x->0

 

Now, you can break / separate them into two:

 

lim   (1-cos2x)   - (cos4x)

x->0       x                x

 

 

As you can see, first one approaches 0 (by looking at second limit above).  Second one should approach 0 as well.

 

You have to use a lot of trig identities to solve this.