Question

limit x tends to zero 1-coss/sin3x into tan2x​

Answer 1

Question :–

$\\ \displaystyle\sf \: \: solve : \lim_{x\:\rightarrow\:0} \left[ \frac{1 - \cos(x) }{ \sin(3x) } \times \tan(2x) \right] \:$

ANSWER :–

$\\ \displaystyle\sf \: \: = \lim_{x\:\rightarrow\:0} \left[ \frac{1 - \cos(x) }{ \sin(3x) } \times \tan(2x) \right] \:$

• We should write this as –

$\\ \displaystyle\sf \: \: = \lim_{x\:\rightarrow\:0} \left[ \frac{1 - \cos(x) }{3x} \times \dfrac{3x}{ \sin(3x) } \times \dfrac{\tan(2x)}{2x} \times 2x \right] \:$

• We know that –

$\\ \displaystyle\sf \: \: \implies { \boxed{ \lim_{x\:\rightarrow\:0} \left[ \frac{ \sin(x) }{x} \right] = 1 \: \: \: and \: \: \lim_{x\:\rightarrow\:0} \left[ \frac{ \tan(x) }{x} \right] = 1 }}$

• So that –

$\\ \displaystyle\sf \: \: = \lim_{x\:\rightarrow\:0} \left[ \frac{1 - \cos(x) }{3 \cancel x} \times 1 \times 1 \times 2 \cancel x \right] \:$

$\\ \displaystyle\sf \: \: = \lim_{x\:\rightarrow\:0} \left[ \frac{1 - \cos(x) }{3 } \times 2 \right] \:$

• Now Applying limits –

$\\ \displaystyle\sf \: \: = \left[ \frac{1 - \cos(0) }{3 } \times 2 \right] \:$

$\\ \displaystyle\sf \: \: = \left[ \frac{1 - 1 }{3 } \times 2 \right] \: \: \: \: \: \: \: \: \: \: \: [ \: \: \because \: \: \: \cos(0) = 1]$

$\\ \displaystyle\sf \: \: = \left[ \frac{0}{3 } \times 2 \right]$

$\\ \displaystyle\sf \: \: = 0$

Hence , $\displaystyle\sf \: \: \lim_{x\:\rightarrow\:0} \left[ \frac{1 - \cos(x) }{ \sin(3x) } \times \tan(2x) \right] \: = 0$