Evaluate $\lim_{x\to 0}\frac{\sqrt[m]{\cos\alpha x}-\sqrt[m]{\cos\beta x}}{\sin^2x},m\in \mathbb{N}$

Question

Evaluate $$\lim_{x\to 0}\frac{\sqrt[m]{\cos\alpha x}-\sqrt[m]{\cos\beta x}}{\sin^2x},m\in \mathbb{N}$$

I used L'Hospital's rule, but that didn't work. Could Taylor series be used? I don't know how to use it with irrational functions.

Answer 1

This can be done via the use of following standard limits $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1},\,\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2},\,\lim_{x \to 0}\frac{\sin x}{x} = 1$$ Let us proceed as follows \begin{align} L &= \lim_{x \to 0}\frac{\sqrt[m]{\cos \alpha x} - \sqrt[m]{\cos \beta x}}{\sin^{2}x}\notag\\ &= \lim_{x \to 0}\frac{\sqrt[m]{\cos \alpha x} - \sqrt[m]{\cos \beta x}}{x^{2}}\cdot\frac{x^{2}}{\sin^{2}x}\notag\\ &= \lim_{x \to 0}\frac{\sqrt[m]{\cos \alpha x} - \sqrt[m]{\cos \beta x}}{x^{2}}\cdot 1\notag\\ &= \lim_{x \to 0}\frac{(1 - \sqrt[m]{\cos \beta x}) - (1 - \sqrt[m]{\cos \alpha x})}{x^{2}}\notag\\ &= \lim_{x \to 0}\frac{(1 - \sqrt[m]{\cos \beta x})}{x^{2}} - \frac{(1 - \sqrt[m]{\cos \alpha x})}{x^{2}}\notag\\ &= \lim_{x \to 0}\frac{1 - \sqrt[m]{\cos \beta x}}{1 - \cos \beta x}\cdot\frac{1 - \cos \beta x}{x^{2}} - \frac{1 - \sqrt[m]{\cos \alpha x}}{1 - \cos \alpha x}\cdot\frac{1 - \cos \alpha x}{x^{2}}\notag\\ &= \lim_{t \to 1}\frac{1 - t^{1/m}}{1 - t}\cdot\lim_{x \to 0}\frac{1 - \cos \beta x}{x^{2}} - \lim_{t \to 1}\frac{1 - t^{1/m}}{1 - t}\cdot\lim_{x \to 0}\frac{1 - \cos \alpha x}{x^{2}}\notag\\ &= \frac{1}{m}\left(\lim_{x \to 0}\frac{1 - \cos \beta x}{x^{2}} - \lim_{x \to 0}\frac{1 - \cos \alpha x}{x^{2}}\right)\notag\\ &= \frac{1}{m}\left(\lim_{x \to 0}\beta^{2}\cdot\frac{1 - \cos (\beta x)}{(\beta x)^{2}} - \lim_{x \to 0}\alpha^{2}\cdot\frac{1 - \cos (\alpha x)}{(\alpha x)^{2}}\right)\notag\\ &= \frac{1}{m}\left(\frac{\beta^{2}}{2} - \frac{\alpha^{2}}{2}\right)\notag\\ &= \frac{\beta^{2} - \alpha^{2}}{2m}\notag \end{align}

Answer 2

You can observe that $a^m-b^m=(a-b)(a^{m-1}+a^{m-2}b+\dots+ab^{m-1}+b^{m-1})$; denote the second factor, for $a=\sqrt[m]{\cos\alpha x}$ and $b=\sqrt[m]{\cos\beta x}$ by $K(x)$.

Then you have $$ \lim_{x\to 0}\frac{\sqrt[m]{\cos\alpha x}-\sqrt[m]{\cos\beta x}}{\sin^2x} =\lim_{x\to0}\frac{\cos\alpha x-\cos\beta x}{\sin^2x\,K(x)} $$ Now observe that $\lim_{x\to0}K(x)=m$ so you're left with $$ \frac{1}{m}\lim_{x\to0}\frac{\cos\alpha x-\cos\beta x}{x^2}\frac{x^2}{\sin^2x} $$ The second factor has limit $1$, so we can disregard it. Then $$ \lim_{x\to0}\frac{\cos\alpha x-\cos\beta x}{x^2}= \lim_{x\to0}\frac{1-(\alpha x)^2/2-1+(\beta x)^2/2+o(x^2)}{x^2} $$

Answer 3

Use \begin{equation} cos(\alpha x)=1-(\alpha x)^{2}/2 \end{equation}

since $x$ is going to zero

\begin{equation} (1-\frac{\alpha^2 x^{2}}{2})^{(1/m)}=1-(\alpha x)^{2}/2m \end{equation}

you should do the same for $(cos(\beta x))^{1/m}$.

Obviously you can replace $sin^2x$ with $x^2$.

Therefore, the final result is: $\frac{1}{2m}(\beta^2-\alpha^2)$.

Answer 4

Using $$\cos x \approx 1 - \dfrac{x^2}{2} + \mathcal{O}(x^4),\quad(1+x)^a \approx 1+ ax + \mathcal{O}(x^2),\quad \sin^2 x \approx x^2 + \mathcal{O}(x^6)$$ It comes : \begin{align} \lim_{x \rightarrow 0} \dfrac{\sqrt[m]{\cos\alpha x}-\sqrt[m]{\cos\beta x}}{\sin^2 x} &= \lim_{x \rightarrow 0}\dfrac{1 + \frac{1}{m} \left( 1 -\frac{(\alpha x)^2}{2}\right) - 1 - \frac{1}{m} \left( 1 - \frac{(\beta x)^2}{2}\right)}{x^2}\\ &= \dfrac{\beta^2 - \alpha^2}{2m} \end{align}

Answer 5

L'Hospital's rule works.$$\lim_{x\to 0}\frac{\sqrt[m]{\cos\alpha x}-\sqrt[m]{\cos\beta x}}{\sin^2 x}$$$$=\lim_{x\to 0}\frac{\frac{1}{m}\cdot (\cos\alpha x)^{\frac{1}{m}-1}\cdot (-\alpha \sin\alpha x)-\frac{1}{m}\cdot (\cos\beta x)^{\frac{1}{m}-1}\cdot (-\beta \sin\beta x)}{2\sin x\cos x}$$$$=\frac{1}{2m}\lim_{x\to 0}\frac{\frac{(\cos\alpha x)^{\frac{1}{m}-1}\cdot (-\alpha \sin\alpha x)}{\alpha\beta x}-\frac{(\cos\beta x)^{\frac{1}{m}-1}\cdot (-\beta \sin\beta x)}{\alpha\beta x}}{\frac{\sin x\cos x}{\alpha\beta x}}$$$$=\frac{1}{2m}\lim_{x\to 0}\frac{\frac{-\alpha(\cos\alpha x)^{\frac 1m-1}}{\beta}\cdot\frac{\sin\alpha x}{\alpha x}-\frac{-\beta (\cos\beta x)^{\frac 1m-1}}{\alpha}\cdot\frac{\sin\beta x}{\beta x}}{\frac{1}{\alpha\beta}\cdot\frac{\sin x}{x}\cdot \cos x}=\frac{-\alpha^2+\beta^2}{2m}$$