Question

Lim x - 0 sin squar 4x / x squar

Answer 1

Question :-

$\to \: \lim_{x \to 0 } \rm \frac{ { \sin}^{2} 4x}{ {x}^{2} }$

Answer :-

$\to \boxed{ \lim_{x \to 0 } \rm \frac{ { \sin}^{2} 4x}{ {x}^{2} } \: = 16} \\ \:$

Solution :-

We have ,

$\to \: \lim_{x \to 0 } \rm \frac{ { \sin} ^{2} 4x}{ {x}^{2} } \\ \: \\ \rm we \: can \: write \: it \: \\ \\ \to \: \lim_{x \to 0 } \rm \frac{ { \sin} 4x}{ {x} } \: \times \frac{ \sin4x}{x} \\ \: \\ \to \:\lim_{x \to 0 } \rm \frac{ { \sin} 4x}{ {x} } \:. \lim_{x \to 0 } \rm \frac{ { \sin} 4x}{ {x} } \: \: \\ \\ \rm multiply \: and \: divided \: by \: 4 \\ \\ \to \: \bigg( \lim_{x \to 0 } \rm \frac{ { \sin} 4x}{ {x} } \: \times \frac{4}{4} \bigg) \bigg( \lim_{x \to 0 } \rm \frac{ { \sin} 4x}{ {x} } \: \times \frac{4}{4} \bigg) \\ \\ \to \bigg( 4\: \: \lim_{x \to 0 } \rm \frac{ { \sin} 4x}{ {4x} } \bigg) \bigg( 4\: \: \lim_{x \to 0 } \rm \frac{ { \sin} 4x}{ {4x} } \bigg) \: \\ \\ \because \: \: \lim_{ } \rm \frac{ { \sin} h}{ {h} } = 1 \\ \\ \to \: 4 \: \times 4 \\ \\ \to \: 16 \\ \to \boxed{ \lim_{x \to 0 } \rm \frac{ { \sin}^{2} 4x}{ {x}^{2} } \: = 16}$