Evaluate the limit.lim x → 0 9 sin(x) − 9xx3

To evaluate 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 given as (9sin(x) - 9x) / x^3.

First, we need to find the derivative of the numerator and the denominator.

The derivative of 9sin(x) is 9cos(x), and the derivative of -9x is -9. So the derivative of the numerator is 9cos(x) - 9.

The derivative of x^3 is 3x^2.

So, according to L'Hopital's Rule, the limit of the function as x approaches 0 is equal to the limit of (9cos(x) - 9) / 3x^2 as x approaches 0.

As x approaches 0, cos(x) approaches 1, so 9cos(x) - 9 approaches 0. And 3x^2 also approaches 0.

So the limit of the function as x approaches 0 is 0/0, which is an indeterminate form. We can apply L'Hopital's Rule again.

The derivative of 9cos(x) - 9 is -9sin(x), and the derivative of 3x^2 is 6x.

So the limit of the function as x approaches 0 is equal to the limit of -9sin(x) / 6x as x approaches 0.

As x approaches 0, sin(x) approaches 0, so -9sin(x) approaches 0. And 6x also approaches 0.

So the limit of the function as x approaches 0 is 0/0 again, which is an indeterminate form. We can apply L'Hopital's Rule once more.

The derivative of -9sin(x) is -9cos(x), and the derivative of 6x is 6.

So the limit of the function as x approaches 0 is equal to the limit of -9cos(x) / 6 as x approaches 0.

As x approaches 0, cos(x) approaches 1, so -9cos(x) approaches -9. And 6 is a constant.

So the limit of the function as x approaches 0 is -9/6, which simplifies to -3/2.

Therefore, the limit of the function as x approaches 0 is -3/2.