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
$\mathop {\lim \limits_{x \to 0} }\frac{98}{83}.\frac{4-cos15x-cos9x-cos5x-cosx}{4.sin^27x}$
$=\mathop {\lim\limits_{x \to 0}} \frac{98}{83}.\frac{\frac{1-cos15x+1-cos9x+1-cos5x+1-cosx}{x^2}}{4.\frac{sin^27x}{x^2}}$
$=\frac{98}{83}.\frac{\frac{15^2}2+\frac{9^2}2+\frac{5^2}2+\frac{1}2}{4.7^2}=1$