Question

Evaluate lim xtends to 0 [1-cos4x]÷[x^2]]

Answer 1

Step-by-step explanation:

$lim_{x \to 0} [1 - cos4x]\div {x}^{2} \\ \\ = lim_{x \to 0}[1 - (1 - 2 {sin}^{2} \: 2x] \div {x}^{2} \\ \\ = lim_{x \to 0}[1 - 1 + 2 {sin}^{2} \: 2x] \div {x}^{2} \\ \\ = lim_{x \to 0}[ 2 {sin}^{2} \: 2x] \div {x}^{2} \\ \\ =2 \times lim_{x \to 0}[ {sin}^{2} \: 2x] \div {x}^{2} \\ \\ =2 \times lim_{x \to 0} \frac{{sin}^{2} \: 2x}{ {x}^{2} } \\ \\ =2 \times lim_{x \to 0}( \frac{{sin} \: 2x}{ {x}}) ^{2} \times \frac{4}{4} \\ \\ =2 \times lim_{x \to 0}( \frac{{sin} \: 2x}{ {2x}}) ^{2} \times 4 \\ \\ =8 \times ( lim_{x \to 0} \: \frac{{sin} \: 2x}{ {2x}}) ^{2} \\ \\ =8 \times( 1) ^2 \\ \\ =8 \times1 \\ \\ = 8 \\ \\ thus \\ \\ \purple{ \boxed{lim_{x \to 0} [1 - cos4x]\div {x}^{2} = 8}}$