$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$L=\frac{a^2m+b^2n}{2mn}$
$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 $\lim_{x\rightarrow 0}\frac{1-\sqrt[n]{\cos bx}}{x^{2}}=\frac{b^2}{2n}$Vậy$L=\frac{a^2+b^2}{2n}$
$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 $\lim_{x\rightarrow 0}\frac{1-\sqrt[
m]{\cos bx}}{x^{2}}=\frac{b^2}{2
m}$Vậy$L=\frac{a^2
m+b^2
n}{2
mn}$