4cos3x.cos5x.cos7x=2cos5x(cos10x+cos4x)=cos15x+cos5x+cos9x+cosxTiếp theo sử dụng định lí: ta tính giới hạn tổng quát sau: Vậy giới hạn cần tìm bằng: limx→09883.4−cos15x−cos9x−cos5x−cosx4.sin27x=limx→09883.1−cos15xx2+1−cos9xx2+1−cos5xx2+1−cosxx2sin27xx2=9883.1522+922+522+1272=9883.8398=1$\mathop {\lim \limits_{x \to 0} }\frac{98}{83}.\frac{4-cos15x-cos9x-cos5x-cosx}{4.sin^27x}$ $=\mathop {\lim \limits_{x \to 0}} \frac{98}{83}.\frac{\frac{1-cos15x}{x^2}+\frac{1-cos9x}{x^2}+\frac{1-cos5x}{x^2}+\frac{1-cosx}{x^2}}{4.\frac{sin^27x}{x^2}}$$=\frac{98}{83}.\frac{\frac{15^2}2+\frac{9^2}2+\frac{5^2}2+\frac{1}2}{4.7^2}=1$
$4
.cos3x.cos5x.cos7x=2
.cos5x
.(cos10x+cos4x)
$$=cos15+cos5x+cos9x+cosx
$Tiếp theo sử dụng định lí:
$\mathop {\lim \limits_{x \to 0} \frac{sinax}a=1}$ ta tính giới hạn tổng quát sau:
$ \mathop {\lim \limits_{x \to 0} \frac{1-cosax}{x^2}=\frac{a^2}2}$ (tự cm nha bạn, ko khó đâu ) Vậy giới hạn cần tìm bằng: limx→09883.4−cos15x−cos9x−cos5x−cosx4.sin27x=limx→09883.1−cos15xx2+1−cos9xx2+1−cos5xx2+1−cosxx2sin27xx2=9883.1522+922+522+1272=9883.8398=1$\mathop {\lim \limits_{x \to 0} }\frac{98}{83}.\frac{4-cos15x-cos9x-cos5x-cosx}{4.sin^27x}$ $=\mathop {\lim \limits_{x \to 0}} \frac{98}{83}.\frac{\frac{1-cos15x}{x^2}+\frac{1-cos9x}{x^2}+\frac{1-cos5x}{x^2}+\frac{1-cosx}{x^2}}{4.\frac{sin^27x}{x^2}}$$=\frac{98}{83}.\frac{\frac{15^2}2+\frac{9^2}2+\frac{5^2}2+\frac{1}2}{4.7^2}=1$