Question-150560

Question Number 150560 by DELETED last updated on 13/Aug/21

Answered by DELETED last updated on 13/Aug/21

\lim_{x \to 3} \frac{(x^{2} - 7 x + 12) \sin (x - 3)}{{(x^{2} - x - 6)}^{2}}

\lim_{x \to 3} \frac{(x - 4) (x - 3) \sin (x - 3)}{{(x + 2) (x - 3)}^{2}}

\lim_{x \to 3} \frac{(x - 4)}{{(x + 2)}^{2} (x - 3)} \times \lim_{(x - 3) \to 0} \frac{\sin (x - 3)}{(x - 3)}

\frac{(3 - 4)}{{(3 + 2)}^{2} (3 - 4)} \times 1 = \frac{(- 1) \times 1}{5^{2} (- 1)} = \frac{- 1}{- 25} = \frac{1}{25}

Commented by tabata last updated on 13/Aug/21

i t h i n g = - \frac{1}{25}

Answered by DELETED last updated on 13/Aug/21

2) . \lim_{x \to - 2} \frac{\tan (6 x + 12)}{(4 x + 8)} = \lim_{(x + 2) \to 0} \frac{\tan 6 (x + 2)}{4 (x + 2)}

\frac{6}{4} \times \lim_{(x + 2) \to 0} \frac{\tan (x + 2)}{(x + 2)} = \frac{6}{4} \times 1 = \frac{3}{2}

Answered by DELETED last updated on 13/Aug/21

3) . \lim_{(x - \frac{π}{8}) \to 0} \frac{\tan (6 x - \frac{3 π}{4})}{\tan (2 x - \frac{π}{4})}

= \lim_{(x - \frac{π}{8}) \to 0} \frac{\tan 6 (x - \frac{π}{8})}{\tan 2 (x - \frac{π}{8})}

= \frac{6}{2} \times \lim_{(x - \frac{π}{8}) \to 0} \frac{\tan (x - \frac{π}{8})}{\tan (x - \frac{π}{8})}

= \frac{6}{2} \times 1 = 3 / /

Facebook Tweet Pin