x lim _(xarrow 0)(1-5x)^(2)/(sinx)=

To find the limit of the given expression as \( x \) approaches 0, we can use L'Hôpital's Rule. L'Hôpital's Rule states that if the limit of a function as \( x \) approaches a point is in the form of \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then the limit of the derivative of the numerator divided by the derivative of the denominator is equal to the original limit.<br /><br />In this case, the given expression is \( \lim_{x \to 0} (1-5x)^{\frac{2}{\sin x}} \). As \( x \) approaches 0, both the numerator and denominator approach 0, so we can apply L'Hôpital's Rule.<br /><br />Taking the natural logarithm of both sides, we get:<br /><br />\[ \ln \left( \lim_{x \to 0} (1-5x)^{\frac{2}{\sin x}} \right) = \lim_{x \to 0} \frac{2}{\sin x} \ln (1-5x) \]<br /><br />Now, we can apply L'Hôpital's Rule to the right-hand side:<br /><br />\[ \lim_{x \to 0} \frac{2}{\sin x} \ln (1-5x) = \lim_{x \to 0} \frac{2 \cdot \frac{d}{dx} \ln (1-5x)}{\frac{d}{dx} \sin x} \]<br /><br />\[ = \lim_{x \to 0} \frac{2 \cdot \frac{-5}{1-5x}}{\cos x} \]<br /><br />\[ = \lim_{x \to 0} \frac{-10}{(1-5x) \cos x} \]<br /><br />As \( x \) approaches 0, the numerator approaches -10 and the denominator approaches 1, so the limit is -10.<br /><br />Now, we can exponentiate both sides to get the original limit:<br /><br />\[ \lim_{x \to 0} (1-5x)^{\frac{2}{\sin x}} = e^{-10} \]<br /><br />Therefore, the limit of the given expression as \( x \) approaches 0 is \( e^{-10} \).