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I'm currently struggling with this task:
$$\lim_{x\to0}{\frac{\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}}}{1-\cos{3x}}}$$
By now I have only come up with the idea to use the formula of sum of cubes, so, the numerator would become $\cos{4x}-\cos{5x}$ and the denominator would be $3(1-\cos{3x})$:
$$\frac{\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}}}{1-\cos{3x}} = \frac{(\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}})*(\sqrt[3]{cos^2{4x}} + \sqrt[3]{\cos{4x}}*\sqrt[3]{\cos{4x}} + \sqrt[3]{\cos^2{5x}})}{(1-\cos{3x})*(\sqrt[3]{cos^2{4x}} + \sqrt[3]{\cos{4x}}*\sqrt[3]{\cos{4x}} + \sqrt[3]{\cos^2{5x}})} = \frac{\cos{4x}-\cos{5x}}{3(1-\cos{3x})}$$ In this expression it looks like $\cos{4x}-\cos{5x}$ and $(1-\cos{3x})$ increase with the same speed (if I can say it this way) and the limit is likely to be $\frac{1}{3}$.
Still I have no idea how to prove that without using L'Hopital's rule which is prohibited by the task.

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  • $\begingroup$ Can't we just use $$\lim_{x\to 0}\cos(4x)-\cos(5x)+\cos(3x)-1=0\implies\lim_{x\to 0}\frac{1}{3}\frac{\cos(4x)-\cos(5x)}{1-\cos(3x)}=\lim_{x\to 0}\frac{1}{3}\frac{1-\cos(3x)}{1-\cos(3x)}=\frac{1}{3}\quad ?$$ $\endgroup$ – Pixel Nov 21 '19 at 13:51
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Once you have rendered

$f(x)=\dfrac{\cos 4x -\cos 5x}{3(1-\cos 3x)}$

you then use the trigonometric sum-product relations to get

$\cos 4x -\cos 5x=2\sin(x/2)\sin(9x/2)$

$1-\cos 3x=\cos 0x - \cos 3x=2\sin^2(3x/2)$

In the latter case, of course, the sum-product relation reduces to a double-angle identity.

Thereby

$f(x)=\dfrac{\sin(x/2)\sin(9x/2)}{3\sin^2(3x/2)}$

Now apply the fact, which probably you have available, that $g(u)=(\sin u)/u\to 1$ as $u\to 0$. Thus

$\sin(x/2)=(x/2)g(x/2)$

and similarly for the other sine arguments. After substitution and cancellation of common factors in the numerator and denominator we end with

$f(x)=\dfrac{g(x/2)g(9x/2)}{3g^2(3x/2)}$

and when we set the $g$ factors to their limit of $1$ we get the overall limit $1/3$.

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  • $\begingroup$ Thanks! That fits me perfectly $\endgroup$ – Daniel Richter Nov 21 '19 at 13:07
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Hint

Use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html

$$\dfrac{\cos2ax-\cos2bx}{1-\cos2cx}=\dfrac{2\sin(b-a)x\cdot\sin(b+a)x}{2\sin^2cx}$$

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You can use the Taylor Expansion of $\cos(x)$.

The numerator is $\dfrac{(5x)^2-(4x)^2}{2!} + O\left({x^4}\right)$

The denominator is $\dfrac{(3x)^2}{3\cdot2!} + O\left({x^4}\right)$

Dividing by $x^2$ and using $3^2+4^2=5^2$ gives the limit as $1/3$.

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  • $\begingroup$ Thanks, it looks like the Taylor Expansion is quite useful here. Unfortunately, it is also prohibited to use, because we haven't learned it yet. Actually, that's why the task is the tricky one $\endgroup$ – Daniel Richter Nov 21 '19 at 11:27
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Adding and subtracting $1$ in the numerator we can rewrite the expression under limit as $$\frac{1-\sqrt [3]{\cos 5x}}{1-\cos 3x}-\frac{1-\sqrt[3]{\cos 4x}} {1-\cos 3x}$$ and the first fraction above can be further rewritten as $$\frac{1-\sqrt[3]{\cos 5x}}{1-\cos 5x}\cdot\frac{1-\cos 5x}{(5x)^2}\cdot\frac{5^2}{3^2}\cdot\frac{(3x)^2}{1-\cos 3x}$$ so that it tends to $(1/3)(1/2)(25/9)(2)=25/27$. Similarly second fraction tends to $16/27$ and hence the desired limit is $9/27=1/3$.

The following standard limits are used in the above approach $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1},\,\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2}$$

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