without-using-lohpital-find-lim-x-pi-6-1-2sinx-cos-3x-

Question Number 31246 by malwaan last updated on 04/Mar/18

without using lohpital

find

\lim_{x \to π / 6} \frac{1 - 2 sinx}{\cos 3 x}

Commented by malwaan last updated on 04/Mar/18

please

I need the answer very soon

Commented by prof Abdo imad last updated on 04/Mar/18

c h x = \frac{π}{6} - t g i v e

{l i m}_{x \to \frac{π}{6}} \frac{1 - 2 s i n x}{c o s (3 x)} = {l i m}_{t \to 0} \frac{1 - 2 s i n (\frac{π}{6} - t)}{c o s (\frac{π}{2} - 3 t)}

= {l i m}_{t \to 0} \frac{1 - 2 (\frac{1}{2} c o s t - \frac{\sqrt{3}}{2} s i n t)}{s i n (3 t)}

= {l i m}_{t \to 0} \frac{1 - c o s t + \sqrt{3} s i n t}{s i n (3 t)} b u t

{l i m}_{t \to 0} \frac{1 - c o s t}{s i n (3 t)} = {l i m}_{t \to 0} \frac{\frac{1 - c o s t}{t}}{\frac{3 s i n (3 t)}{3 t}} = \frac{0}{3} = 0 a n d

{l i m}_{t \to 0} \frac{\sqrt{3} s i n t}{s i n (3 t)} = {l i m}_{t \to 0} \sqrt{3} \frac{\frac{s i n t}{t}}{\frac{3 s i n (3 t)}{3 t}} = \frac{\sqrt{3}}{3} .

Answered by naka3546 last updated on 04/Mar/18

Commented by malwaan last updated on 05/Mar/18

thank you

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