Question

limit x tends to 0 (sin^2ax)/(sin^2bx)​

Answer 1

Given Expression,

$\displaystyle \sf \lim_{x \to \: 0} \: \dfrac{ {sin}^{2}ax }{ {sin}^{2} bx}$

Multiplying and dividing by (ax)² in numerator and (bx)² in denominator,

$\implies \displaystyle \sf \lim_{x \to \: 0} \: \dfrac{(ax) {}^{2} {sin}^{2}ax }{(ax) {}^{2} } \times \dfrac{(bx) {}^{2} }{ {sin}^{2}bx(bx) {}^{2} } \\ \\ \implies \displaystyle \sf {a}^{2} {x}^{2} \lim_{x \to \: 0} \: \dfrac{ {sin}^{2}ax }{(ax) {}^{2} } \times \dfrac{1}{ {b}^{2} {x}^{2} } \lim_{x \to \: 0}\dfrac{(bx) {}^{2} }{ {sin}^{2}bx} \\ \\ \implies \displaystyle \sf {a}^{2} {x}^{2} \lim_{x \to \: 0} \: \dfrac{ {sin \: }^{}ax }{(ax) {}^{} } \times\lim_{x \to \: 0} \: \dfrac{ {sin \: }^{}ax }{(ax) {}^{} } \times \dfrac{1}{ {b}^{2} {x}^{2} } \lim_{x \to \: 0}\dfrac{(bx) {}^{} }{ {sin}^{ \: }bx} \times \lim_{x \to \: 0}\dfrac{(bx) {}^{} }{ {sin}^{ \: }bx} \\ \\ \implies \sf \: \dfrac{a {}^{2} { {x}^{2}} }{ {b}^{2} {x}^{2} } \times 1 \\ \\ \implies \sf \: \dfrac{a {}^{2} { } }{ {b}^{2} }$

The value of the above expression at x tends to 0 is a²/b².