Lim x - 0sin 3 -tan 2 -

Lim_x→ 0sin 3θ/tan 2θ =lim_0→ 0sin 3θ/3θ× 3θ/lim_20→ 0tan 2θ/2θ× 2θ = (lim_30→ 0sin 3θ/3θ )/ ( lim_20→ 0tan 2θ/2θ )×3θ/2θ =1/2×3/2 [ ∵lim_x→ 0sinx/ ...

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Byju's Answer

Standard XII

Mathematics

Sandwich Theorem

lim x → 0sin ...

Question

${lim}_{x \to 0} \frac{s i n 3 θ}{t a n 2 θ}$

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Solution

${lim}_{x \to 0} \frac{s i n 3 θ}{t a n 2 θ}$

$= \frac{{lim}_{0 \to 0} \frac{s i n 3 θ}{3 θ} \times 3 θ}{{lim}_{20 \to 0} \frac{t a n 2 θ}{2 θ} \times 2 θ}$

$= \frac{({lim}_{30 \to 0} \frac{s i n 3 θ}{3 θ})}{({lim}_{20 \to 0} \frac{t a n 2 θ}{2 θ})} \times \frac{3 θ}{2 θ}$

$= \frac{1}{2} \times \frac{3}{2}$

$[∵ {lim}_{x \to 0} \frac{s i n x}{x} = 1 and {lim}_{x \to 0} \frac{t a n x}{x} = 1]$

$= \frac{3}{2}$

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Similar questions

Prove:

[\frac{1 + {tan}^{2} θ}{1 + {cot}^{2} θ}] = {[\frac{1 - tan θ}{1 - cot θ}]}^{2} = {tan}^{2} θ

Q. If

θ = 30^{\circ}

, verify that :
(i)

tan 2 θ = \frac{2 tan θ}{1 - {tan}^{2} θ}

(ii)

tan 2 θ = \frac{4 tan θ}{1 + {tan}^{2} θ}

(iii)

cos 2 θ = \frac{1 - {tan}^{2} θ}{1 + {tan}^{2} θ}

(iv)

cos 3 θ = 4 {cos}^{3} θ - 3 cos θ

\frac{{cos}^{2} θ + {tan}^{2} θ - 1}{{sin}^{2} θ} = {tan}^{2} θ

Q. Solve:

\frac{1 + t a n^{2} θ}{1 - t a n^{2} θ} = s e c 2 θ .

$\frac{t a n^{2} θ}{1 + t a n^{2} θ}$ + $\frac{c o t^{2} θ}{1 + c o t^{2} θ}$ =

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limx→0sin 3θtan 2θ

limx→0sin 3θtan 2θ =lim0→0sin 3θ3θ×3θlim20→0tan 2θ2θ×2θ =(lim30→0sin 3θ3θ)(lim20→0tan 2θ2θ)×3θ2θ =12×32 [∵limx→0sinxx=1 andlimx→0tanxx=1] =32

${lim}_{x \to 0} \frac{s i n 3 θ}{t a n 2 θ}$