Question

Prove it. The topic is Limits and Continuity.

Answer 1

Topic :-

Limits

To Prove :-

$\displaystyle \lim_{x \to 0}\tt {\:\:\dfrac{sin^3a\theta}{\theta sin^2b\theta}=\dfrac{a^3}{b^2}}$

Formula to be Used :-

$\displaystyle \lim_{x \to 0}\tt {\:\:\dfrac{sinx}{x}=1}$

Proof :-

Solving LHS,

$\displaystyle \lim_{x \to 0}\tt {\:\:\dfrac{sin^3a\theta}{\theta sin^2b\theta}}$

$\displaystyle \lim_{x \to 0}\tt {\:\:\dfrac{(sina\theta)^3}{\theta (sinb\theta)^2}}$

We can write it as,

$\displaystyle \lim_{x \to 0}\tt {\:\:\left ( \dfrac{a^3\theta^3}{b^2\theta^2} \right )\left ( \dfrac{b^2\theta^2}{a^3\theta^3} \right )\dfrac{(sina\theta)^3}{(sinb\theta)^2}\dfrac{1}{\theta}}$

$\displaystyle \lim_{x \to 0}\tt {\:\:\left ( \dfrac{a^3\theta^3}{b^2\theta^2} \right )\left ( \dfrac{sina\theta}{a\theta} \right )^3\left ( \dfrac{b\theta}{sinb\theta} \right )^2\dfrac{1}{\theta}}$

Applying formula gives,

$\tt {\left ( \dfrac{a^3\theta}{b^2} \right )\left ( 1 \right )^3\left ( 1 \right )^2\left ( \dfrac{1}{\theta} \right )}$

$\tt {\left ( \dfrac{a^3\theta}{b^2\theta} \right )}$

$\tt {\left ( \dfrac{a^3}{b^2} \right )}$

RHS is given as,

$\tt {\left ( \dfrac{a^3}{b^2} \right )}$

LHS = RHS

Hence, Proved!