Nilai \( \displaystyle \lim_{x\to 0} \ \frac{\sin (4x^2)}{x^2+\tan^2 3x} = \cdots \)
- 1/5
- 2/5
- 3/5
- 4/5
- 1
Pembahasan:
\begin{aligned} \lim_{x\to 0} \ \frac{\sin (4x^2)}{x^2+\tan^2 3x} &= \lim_{x\to 0} \ \frac{\sin (4x^2)}{x^2+\tan^2 3x} \times \frac{\frac{1}{x^2}}{\frac{1}{x^2}} \\[8pt] &= \lim_{x\to 0} \ \frac{ \frac{\sin (4x^2)}{x^2}}{1+\frac{\tan^2 3x}{x^2}} \\[8pt] &= \frac{\displaystyle \lim_{x\to 0} \ \frac{\sin (4x^2)}{x^2}}{\displaystyle \lim_{x\to 0} \ 1 + \lim_{x\to 0} \ \frac{\tan^2 3x}{x^2}} \\[8pt] &= \frac{4}{1+3^3} = \frac{4}{10}= \frac{2}{5} \end{aligned}
Jawaban B.