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

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