Compute: $$\lim_{x\to 0}\frac{\sqrt[m]{\cos(\alpha
x)}-\sqrt[n]{\cos{(\beta x)}}}{x^2},\;m,n\in\mathbb N,\alpha,\beta\in\mathbb R$$
My attempt:
$$L=\lim_{x\to 0}\frac{\sqrt[m]{\cos(\alpha x)}-\sqrt[n]{\cos{(\beta x)}}}{x^2}=\lim_{x\to 0}\frac{\sqrt[m]{\cos(\alpha x)}-1+1-\sqrt[n]{\cos(\beta x)}}{x^2}\\\displaystyle=\lim_{x\to 0}\frac{\left(1+(\cos(\alpha x)-1)\right)^{\frac{1}{m}}-1-\left(\left(1+(\cos(\beta x)-1)\right)^{\frac{1}{n}}-1\right)}{x^2}$$
Transformed expression:
$$-\frac{(1+(\cos(\alpha x)-1))^{\frac{1}{m}}-1}{\cos(\alpha x)-1}\cdot\frac{1-\cos(\alpha x)}{\alpha^2x^2}\alpha^2+\frac{(1+(\cos(\beta x)-1))^{\frac{1}{n}}-1}{\cos(\beta x)-1}\cdot\frac{1-\cos(\beta x)}{\beta^2x^2}\beta^2$$
I applied the standard limits from the table: $$\lim_{x\to 0}\frac{(1+x)^a-1}{x}=a\;\&\;\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$
$$L=-\frac{1}{m}\cdot\frac{1}{2}\alpha^2+\frac{1}{n}\cdot\frac{1}{2}\beta^2=\frac{m\beta^2-n\alpha^2}{2mn}$$
Is this correct?
Best Answer
Your initial split by adding subtracting $1$ is fine. The next step requires the standard limit for algebraic functions namely $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ The above is simpler than using $((1+x)^a-1)/x\to a$. Thus you can write the term $$\frac{1-\sqrt[n]{\cos \beta x}} {x^2}$$ as $$\frac{1-(\cos\beta x) ^{1/n}}{1-\cos \beta x} \cdot\frac{1-\cos\beta x} {(\beta x) ^2}\cdot\beta^2$$ which tends to $$\frac{1}{n}\cdot\frac{1}{2}\cdot\beta^2$$ Now you know that the final answer is $$\frac{m\beta^2-n\alpha^2} {2mn}$$