Compute   lim⁡ 𝑥→0⁡𝑠𝑖𝑛2𝑥𝑠𝑖𝑛5𝑥

To compute the limit of the function as x approaches 0, we can use L'Hopital's Rule. This rule states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives.

The function is sin^2(x)/sin^5(x).

First, let's simplify this function: sin^2(x)/sin^5(x) = 1/sin^3(x).

Now, let's find the derivative of the numerator and the denominator.

The derivative of 1 (numerator) is 0.

The derivative of sin^3(x) (denominator) is 3sin^2(x)cos(x).

So, according to L'Hopital's Rule, the limit of our function as x approaches 0 is the limit of the quotient of these derivatives as x approaches 0.

That is, lim (x->0) [0/(3sin^2(x)cos(x))].

Since the numerator is 0, the limit of the whole function as x approaches 0 is 0.