Kindly consider the following limθ→01−cos4θ1−cos6θ is (a) 4 9 (b) 1 2 (c) − 1 2 (d) -1

This is an objective Type Questions as classified in NCERT Exemplar Sol: Given that limθ→01−cos4θ1−cos6θ =limθ→02sin22θ2sin23θ [?1−cosθ=2sin2θ2] =limθ→0sin22θsin23θ=limθ→0[sin2θsin3θ]2=limθ→02θ→03θ→0[sin2θ2θ*2θsin3θ3θ*3θ]2 =[2θ3θ]2=(23)2=49Hence, the correct option is (a).

Kindly consider the following

$l i m_{θ \to 0} \frac{1 - c o s 4 θ}{1 - c o s 6 θ}$ is

(a) $\frac{4}{9}$
(b) $\frac{1}{2}$
(c) $- \frac{1}{2}$
(d) -1

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Answered by

Alok Kumar Singh | Contributor-Level 10

7 months ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{θ \to 0}{l i m} \frac{1 - c o s 4 θ}{1 - c o s 6 θ} \\ = \underset{θ \to 0}{l i m} \frac{2 s i n^{2} 2 θ}{2 s i n^{2} 3 θ} [? 1 - c o s θ = 2 s i n^{2} \frac{θ}{2}] \\ = \underset{θ \to 0}{l i m} \frac{s i n^{2} 2 θ}{s i n^{2} 3 θ} = \underset{θ \to 0}{l i m} {[\frac{s i n 2 θ}{s i n 3 θ}]}^{2} = \underset{\begin{array}{l} θ \to 0 \\ 2 θ \to 0 \\ 3 θ \to 0 \end{array}}{l i m} {[\frac{\frac{s i n 2 θ}{2 θ} * 2 θ}{\frac{s i n 3 θ}{3 θ} * 3 θ}]}^{2} \\ = {[\frac{2 θ}{3 θ}]}^{2} = {(\frac{2}{3})}^{2} = \frac{4}{9} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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