Evaluating the Limit of cos(3x)-cos(4x)/x^2: The Squeeze Theorem

[tylersmith7690]

Homework Statement

evaluate the following limit.

lim_x→∞ cos3x-cos4x/x^2 , include theorems

Homework Equations

Im guessing its a sandwhich theorem limit and that cos3x-cos4x as an upper bound of 3 and lower of -3. But was wondering if anyone can help explain to me why this is so and should I show a proof or working for this assumption.

The Attempt at a Solution

-3 ≤ cos 3x - cos 4x ≤ 3 , divide through by x^2 to get original equation

-3/x^2 ≤ ( cos 3x - cos 4x ) / x²≤ 4/x²

Now lim_x→∞ -3/x² = 0 and
lim_x→∞ 3/x² = 0.

So the limx→∞ ( cos 3x - cos 4x ) / x² = 0 by the squeeze theorem.

I'm not sure as to what rules I would have to put in this working out. Except maybe something

about -1≤ cos ≤ 1 and how cos(3x) = cos(2x + x) = 3cos³-cos x

and cos(4x) = 2 cos2(2x) - 1 = 2(2 cos2(x) - 1)2 - 1 = 8 cos4(x) - 8 cos2(x) + 1
but i don't see how this would help.

 

[ehild]

Homework Statement

evaluate the following limit.

lim_x→∞ cos3x-cos4x/x^2 , include theorems

lim_x→∞ cos3x-cos4x/x^2 does not exist. Unless you intended to write lim_x→∞ (cos3x-cos4x)/x^2?

2. Homework Equations

Im guessing its a sandwhich theorem limit and that cos3x-cos4x as an upper bound of 3 and lower of -3. But was wondering if anyone can help explain to me why this is so and should I show a proof or working for this assumption.

I do not see why 3 is chosen as upper bound and -3 as the lower one. But they are valid bounds. Anyway, -1≤cos3x≤1, and the same is true for cos(4x), -1≤cos4x≤1. It can happen, that cos 3x=-1 and at the same time cos 4x= 1, or cos(3x)=1 and cos(4x)=-1, So -2≤cos(3x)-cos(4x) ≤2 is always true, and you can apply the sandwich method.

The Attempt at a Solution

-3 ≤ cos 3x - cos 4x ≤ 3 , divide through by x^2 to get original equation

-3/x^2 ≤ ( cos 3x - cos 4x ) / x²≤ 4/x²

Now lim_x→∞ -3/x² = 0 and
lim_x→∞ 3/x² = 0.

So the limx→∞ ( cos 3x - cos 4x ) / x² = 0 by the squeeze theorem..

That is right, the limit is zero.

I'm not sure as to what rules I would have to put in this working out. Except maybe something

about -1≤ cos ≤ 1 and how cos(3x) = cos(2x + x) = 3cos³-cos x

and cos(4x) = 2 cos2(2x) - 1 = 2(2 cos2(x) - 1)2 - 1 = 8 cos4(x) - 8 cos2(x) + 1
but i don't see how this would help.

No reason to complicate it... ehild

 

[vanhees71]

I would use Taylor expansion of the numerator
\cos(3 x)-\cos(4 x)=1-\frac{9 x^2}{2}-1+\frac{(4 x)^2}{2} + \mathcal{O}(x^4) = \frac{7}{2} x^2 + \mathcal{O}(x^4)
and then I'd think about the limit again (it's NOT 0!).

BTW: It is good to use LaTeX for the formulae. The OP was not well defined and I could only guess what the right expression should be!

 

[ehild]

I would use Taylor expansion of the numerator
\cos(3 x)-\cos(4 x)=1-\frac{9 x^2}{2}-1+\frac{(4 x)^2}{2} + \mathcal{O}(x^4) = \frac{7}{2} x^2 + \mathcal{O}(x^4)
and then I'd think about the limit again (it's NOT 0!).

The Taylor expansion does not work for x→∞.

ehild