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$L=\lim_{x\rightarrow 0}\frac{1-\sqrt[m]{\cos bx}.\sqrt[n]{\cos ax}}{x^{2}}$
$=\lim_{x\rightarrow 0}\frac{1-\sqrt[n]{\cos ax} +(1-\sqrt[m]{\cos bx})\sqrt[n]{\cos ax}}{x^{2}}$
$=\lim_{x\rightarrow 0}\frac{1-\sqrt[n]{\cos ax}}{x^{2}}+\lim_{x\rightarrow 0}\frac{(1-\sqrt[m]{\cos bx})\sqrt[n]{\cos ax}}{x^{2}}$
$=\lim_{x\rightarrow 0}\frac{1-\sqrt[n]{\cos
ax}}{x^{2}}+\lim_{x\rightarrow 0}\frac{1-\sqrt[m]{\cos
bx}}{x^{2}}$, do $\lim_{x\rightarrow 0} \cos ax =1$
ta tính
$\lim_{x\rightarrow 0}\frac{1-\sqrt[n]{\cos
ax}}{x^{2}}=\lim_{x\rightarrow 0}\frac{1-\cos
ax}{x^{2}}.\frac{1}{1+\sqrt[n]{\cos
ax}+\cdots+\sqrt[n]{(\cos
ax)^{n-1}}}=\lim_{x\rightarrow 0}\frac{2\sin^2\frac{ax}{2}}{x^2}\frac{1}{n}$
$=\lim_{x\rightarrow 0}\frac{a^2}{2}\left ( \frac{\sin \frac{ax}{2}}{\frac{ax}{2}}\right )^2\frac{1}{n}=\frac{a^2}{2n}$
Tương tự ta cũng có
$\lim_{x\rightarrow 0}\frac{1-\sqrt[m]{\cos bx}}{x^{2}}=\frac{b^2}{2m}$
Vậy
$L=\frac{a^2m+b^2n}{2mn}$
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