limit x->0 (tan^2 (1/6)x)/(sin^2 (1/2)x)=...

1/9

Pembahasan

Untuk menyelesaikan limit $\lim_{x\to 0} \frac{\tan^2 (\frac{1}{6}x)}{\sin^2 (\frac{1}{2}x)}$, kita dapat menggunakan sifat limit $\lim_{u\to 0} \frac{\tan u}{u} = 1$ dan $\lim_{u\to 0} \frac{\sin u}{u} = 1$.
Kita bisa menulis ulang persamaan menjadi:

$\lim_{x\to 0} \left(\frac{\tan (\frac{1}{6}x)}{\frac{1}{6}x}\right)^2 \cdot \left(\frac{\frac{1}{2}x}{\sin (\frac{1}{2}x)}\right)^2 \cdot \frac{(\frac{1}{6}x)^2}{(\frac{1}{2}x)^2}$

Karena $\lim_{x\to 0} \frac{\tan (\frac{1}{6}x)}{\frac{1}{6}x} = 1$ dan $\lim_{x\to 0} \frac{\sin (\frac{1}{2}x)}{\frac{1}{2}x} = 1$, maka:

$1^2 \cdot 1^2 \cdot \frac{(\frac{1}{6})^2}{(\frac{1}{2})^2} = \frac{\frac{1}{36}}{\frac{1}{4}} = \frac{1}{36} \cdot 4 = \frac{4}{36} = \frac{1}{9}$.

Jadi, $\lim_{x\to 0} \frac{\tan^2 (\frac{1}{6}x)}{\sin^2 (\frac{1}{2}x)} = \frac{1}{9}$.