lim _(x arrow 0) (1+cos 5 x)/(sin ^2) 3 x=

To find the limit \( \lim_{x \rightarrow 0} \frac{1+\cos 5x}{\sin^2 3x} \), we can use L'Hôpital's Rule since both the numerator and denominator approach 0 as \( x \) approaches 0.<br /><br />First, let's rewrite the expression:<br />\[ \lim_{x \rightarrow 0} \frac{1+\cos 5x}{\sin^2 3x} \]<br /><br />We can apply L'Hôpital's Rule, which states that if \(\lim_{x \to c} \frac{f(x)}{g(x)}\) results in an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then:<br />\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]<br /><br />First, we need to find the derivatives of the numerator and the denominator.<br /><br />The numerator is \(1 + \cos 5x\). Its derivative is:<br />\[ \frac{d}{dx}(1 + \cos 5x) = -5 \sin 5x \]<br /><br />The denominator is \(\sin^2 3x\). Using the chain rule, its derivative is:<br />\[ \frac{d}{dx}(\sin^2 3x) = 2 \sin 3x \cdot \cos 3x \cdot 3 = 6 \sin 3x \cos 3x \]<br /><br />Now, applying L'Hôpital's Rule:<br />\[ \lim_{x \rightarrow 0} \frac{1+\cos 5x}{\sin^2 3x} = \lim_{x \rightarrow 0} \frac{-5 \sin 5x}{6 \sin 3x \cos 3x} \]<br /><br />As \( x \) approaches 0, \(\sin 5x\) and \(\sin 3x\) also approach 0. We can apply L'Hôpital's Rule again:<br /><br />The derivative of \(-5 \sin 5x\) is:<br />\[ \frac{d}{dx}(-5 \sin 5x) = -25 \cos 5x \]<br /><br />The derivative of \(6 \sin 3x \cos 3x\) is:<br />\[ \frac{d}{dx}(6 \sin 3x \cos 3x) = 6 (\cos 3x \cdot \cos 3x - \sin 3x \cdot \sin 3x) \cdot 3 = 18 (\cos^2 3x - \sin^2 3x) \]<br /><br />So, applying L'Hôpital's Rule again:<br />\[ \lim_{x \rightarrow 0} \frac{-5 \sin 5x}{6 \sin 3x \cos 3x} = \lim_{x \rightarrow 0} \frac{-25 \cos 5x}{18 (\cos^2 3x - \sin^2 3x)} \]<br /><br />As \( x \) approaches 0, \(\cos 5x\) approaches 1 and \(\cos^2 3x - \sin^2 3x\) approaches 1:<br />\[ \lim_{x \rightarrow 0} \frac{-25 \cos 5x}{18 (\cos^2 3x - \sin^2 3x)} = \frac{-25 \cdot 1}{18 \cdot 1} = \frac{-25}{18} \]<br /><br />Therefore, the limit is:<br />\[ \lim_{x \rightarrow 0} \frac{1+\cos 5x}{\sin^2 3x} = \frac{-25}{18} \]