Question

1-COS4x
Q.10 limx-0
x2​

Answer 1

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\large \boxed {\red{\tt :  ⟼1 - cosx = 2 {sin}^{2} \dfrac{x}{2} }}$

$\large \boxed {\red{\tt :  ⟼ \: \lim_{x\to 0} \: \dfrac{sinx}{x} \: = 1}}$

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$\large\underline\purple{\bold{Solution :- }}$

$\tt :  ⟼ \: \lim_{x\to 0} \: \dfrac{1 - cos4x}{ {x}^{2} }$

☆On substituting directly x = 0, we get indeterminant form

$\tt :  ⟼ \: \lim_{x\to 0} \: \dfrac{2 \: {sin}^{2}2x }{ {x}^{2} }$

$\tt :  ⟼ \: 2 \: \lim_{x\to 0} \: \dfrac{sin2x \: \times sin2x}{ {x}^{2} }$

$\tt :  ⟼ \: 2 \: \lim_{x\to 0} \: \dfrac{sin2x}{x} \times \lim_{x\to 0}\dfrac{sin2x}{x}$

$\tt :  ⟼ \: 2 \: \lim_{x\to 0} \: \dfrac{sin2x}{2x} \times 2 \times \lim_{x\to 0}\dfrac{sin2x}{2x} \times 2$

$\tt :  ⟼ \: 2 \times 1 \times 2 \times 1 \times 2$

$\tt :  ⟼ \: 8$

$\large \boxed {\red{\tt :  ⟼ \: Hence \: \lim_{x\to 0}\dfrac{1 - cos4x}{ {x}^{2} } = 8}}$