Soal dan Penyelesaian Limit Trigonometri

Soal 1:

\(lim_{ x\to 0}\frac{sin3x-sin3x.cos2x}{2x^3}= ...\)

Penyelesaian:

\(\begin{array}{rcl} lim_{ x\to 0}\frac{sin3x-sin3x.cos2x}{2x^3}&=&lim_{ x\to 0}\frac{sin3x\left ( 1-cos2x \right )}{2x^3}\\&=&lim_{ x\to 0}\frac{sin3x\left ( 2sin^2x \right )}{2x^3} \\&=&lim_{ x\to 0}\frac{2sin3x.sin^2x}{2x^3} \\&=&lim_{ x\to 0}\frac{2sin3x}{2x} . \frac{sinx}{x}. \frac{sinx}{x}\\&=&\frac{2.3.1.1}{2.1.1.1}\\&=&3\\ \end{array}\)

Soal 2:

\(lim_{ x\to 0}\frac{cos4x+sin2x.sin4x-1}{4x^2.sin^2x}=...\)

Penyelesaian:

\(\begin{array}{rcl} lim_{ x\to 0}\frac{cos4x+sin2x.sin4x-1}{4x^2.sin^2x}&=& lim_{ x\to 0}\frac{cos4x-1+sin2x.sin4x}{4x^2.sin^2x}\\ &=&lim_{ x\to 0}\frac{-2sin^22x+sin2x.2sin2x.cos2x}{4x^2.sin^2x}\\ &=&lim_{ x\to 0}\frac{-2sin^22x+2sin^22x.cos2x}{4x^2.sin^2x}\\ &=&lim_{ x\to 0}\frac{-2sin^22x\left ( 1-cos2x \right )}{4x^2.sin^2x}\\ &=&lim_{ x\to 0}\frac{-2sin^22x\left ( 2sin^2x \right )}{4x^2.sin^2x}\\ &=&lim_{ x\to 0}\frac{-4sin^22x.sin^2x}{4x^2.sin^2x}\\&=&lim_{ x\to 0}\frac{-4sin^22x}{4x^2}.\frac{sin^2x}{sin^2x}\\ &=&\frac{-4.2^2}{4.1^2}.\frac{1^2}{1^2}\\ &=&-4\\ \end{array} \)