SURENDRA K. answered • 09/22/14

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Sarah J.

asked • 09/22/14fin the limit of (1-cos2x-cos4x)/x as x->0 and show all work!

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SURENDRA K. answered • 09/22/14

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An experienced,patient & hardworking tutor

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

Nadia N. answered • 09/22/14

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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.

Yohan C. answered • 09/22/14

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Math Tutor (up to Calculus) (not Statistics and Finite)

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.

Yohan C.

I didn't calculate second part (cos 4x / x). Here is how I did it and it came out to be ∞ or doesn't exist.

lim cos 4x = 4 cos 4x = 4 lim cos 4x = 4 lim cos u = 4 lim [ ln cos u]

x->0 x 4x 4x u [ u ]

ln cos u becomes ln 1 which is 0 but [ -ln 1/u] or ln u or ln 0 which it gives undefined value (∞). If this limit is approaching 0^{+} from the right, -∞ might have been the answer. But since it approaches 0, this limit doesn't exists.

Sorry for mishap.

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09/23/14

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Yohan C.

09/23/14