lim _(xarrow 0)(sin^25x)/(4x^2)

To find the limit of the given expression as \( x \) approaches 0, we can use the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).<br /><br />First, rewrite the given expression:<br /><br />\[<br />\lim_{x \to 0} \frac{\sin^2(5x)}{4x^2}<br />\]<br /><br />We can split this into two parts:<br /><br />\[<br />\lim_{x \to 0} \frac{1}{4} \cdot \frac{\sin^2(5x)}{x^2}<br />\]<br /><br />Next, use the identity \( \sin^2(5x) = (\sin(5x))^2 \):<br /><br />\[<br />\lim_{x \to 0} \frac{1}{4} \cdot \left( \frac{\sin(5x)}{x} \right)^2<br />\]<br /><br />Now, apply the standard limit \( \lim_{x \to 0} \frac{\sin(5x)}{x} = 5 \):<br /><br />\[<br />\lim_{x \to 0} \frac{1}{4} \cdot (5)^2 = \frac{1}{4} \cdot 25 = \frac{25}{4}<br />\]<br /><br />Therefore, the limit is:<br /><br />\[<br />\boxed{\frac{25}{4}}<br />\]