lim _{x rightarrow 0} frac{x^2 sin ^2(3 x)+si | vedclass

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Question

(A) We know that $\lim _{x \rightarrow 0} \frac{\sin(ax)}{ax} = 1$ and $1 - \cos(ax) = 2 \sin^2(\frac{ax}{2})$.Given the expression: $\lim _{x \righta

Vedclass · Accepted Answer

$\frac{580}{9}$

Vedclass · Answer

(A) We know that $\lim _{x ightarrow 0} \frac{\sin(ax)}{ax} = 1$ and $1 - \cos(ax) = 2 \sin^2(\frac{ax}{2})$.Given the expression: $\lim _{x ightarrow 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{(1-\cos 3 x)^2}$.Using the identity $1 - \cos(3x) = 2 \sin^2(\frac{3x}{2})$,the denominator becomes $(2 \sin^2(\frac{3x}{2}))^2 = 4 \sin^4(\frac{3x}{2})$.Now,divide numerator and denominator by $x^4$:Numerator: $\frac{x^2 \sin^2(3x)}{x^4} + \frac{\sin^4(6x)}{x^4} = \frac{\sin^2(3x)}{x^2} + \frac{\sin^4(6x)}{x^4}$.As $x ightarrow 0$,$\frac{\sin^2(3x)}{x^2} ightarrow (3)^2 = 9$ and $\frac{\sin^4(6x)}{x^4} ightarrow (6)^4 = 1296$.Denominator: $\frac{4 \sin^4(\frac{3x}{2})}{x^4} = 4 \cdot (\frac{\sin(\frac{3x}{2})}{\frac{3x}{2}})^4 \cdot (\frac{3}{2})^4 = 4 \cdot 1 \cdot \frac{81}{16} = \frac{81}{4}$.Thus,the limit is $\frac{9 + 1296}{81/4} = \frac{1305 	imes 4}{81} = \frac{145 	imes 4}{9} = \frac{580}{9}$.

$\lim _{x \rightarrow 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{(1-\cos 3 x)^2}=$

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