Rumus Limit Fungsi Trigonometri

\[lim_{x \to 0} \frac{sin \ x}{x} = lim_{x \to 0} \frac{x}{sin \ x} = 1\] \[lim_{x \to 0} \frac{tan \ x}{x} = lim_{x \to 0} \frac{x}{tan \ x} = 1\] \[lim_{x \to 0} \frac{sin \ ax}{bx} = lim_{x \to 0} \frac{ax}{sin \ bx} = \frac{a}{b}\] \[lim_{x \to 0} \frac{tan \ ax}{bx} = lim_{x \to 0} \frac{ax}{tan \ bx} = \frac{a}{b}\] \[lim_{x \to 0} \frac{sin \ ax}{sin \ bx} = lim_{x \to 0} \frac{tan \ ax}{tan \ bx} = \frac{a}{b}\] \[lim_{x \to 0} \frac{sin \ ax}{tan \ bx} = lim_{x \to 0} \frac{tan \ ax}{sin \ bx} = \frac{a}{b}\] \[lim_{x \to 0} \frac{sin \ ax +tan \ bx}{cx – sin \ dx} = \frac{a+b}{c-d}\] \[lim_{x \to 0} \frac{1 -cos \ x}{x} = 0\]

\[lim_{x \to 1} = \frac{(x^2 -1) + tan (2x -2)}{sin^2 (x-1)}\]

\[lim_{x \to 1} = \frac{(x^2 -1) + tan (2x -2)}{sin^2 (x-1)}\] \[lim_{x \to 1} = ( \frac {(x-1)}{sin (x-1)} \cdot (x+1) \cdot \frac{tan \ 2(x-1)}{sin (x-1)})\] \[1 \cdot lim_{x \to 1} (x+1) \cdot 2\] \[=4\]