 We can use the conjugate to help find limits involving radicals. Remember, the conjugate of an expression involving a square root has the same terms, but the operation, whether it's addition or subtraction, is switched. And a useful idea, factored form, is best. So unless you need to expand a product, don't bother. So let's try to find this limit. At x equals 9, the expression has indeterminate form 0 over 0, so we have to do some algebra. Since this has a square root, we'll multiply numerator and denominator by a conjugate. And since square root of x minus 3 is a subtraction, the conjugate will have the same terms added. So we'll take our expression, multiply numerator and denominator by the conjugate. Now remember, the whole purpose of the conjugate is that the product of an expression and its conjugate will eliminate the square root. This means we should expand the product in the numerator to get. However, we know nothing about the product of a conjugate and the arbitrary polynomial, so until and unless we need to expand it, we'll leave the denominator in product form. And we see we do have this common factor of x minus 9, so we'll remove it and get. And since our new expression is algebraic and defined at x equals 9, then the limit is the function value at x equals 9, which will be... If we have an expression with multiple square roots, remember we can take them one at a time. So note that at x equals 1, both our numerator and denominator are 0, so this expression has indeterminate form 0 over 0, so we can do some algebra. We'll eliminate the square root in the denominator by multiplying by its conjugate. So again, those will have the same terms, but since the expression is subtracted, we'll add them. And we have to multiply numerator by the same expression. And again, there's no reason to expand the product in the numerator, so we'll leave it in factored form. But expanding the denominator gives us... And we'll simplify this a little bit. One of the reasons we want to leave this in factored form is our numerator still has this expression square root 3x plus 1 minus 2, so it's still 0 at x equals 1. Unfortunately, the denominator is also still 0 at x equals 1, and so there's more algebra to do. So, let's multiply by the conjugate of the original numerator. And so this will be the same terms with an addition instead of a subtraction. And again, we only care about the product of the conjugate expressions, so we'll leave the others in factored form, meanwhile we should multiply the expression at its conjugate. But if we do that, we get and simplify. And we note that x equals 1 makes 3x minus 3 and 24 minus 24x equal to 0. So again, our factor theorem says these both have a factor of x minus 1. So 3x minus 3 must be x minus 1 times 3, and 24 minus 24x must be x minus 1 times negative 24. And again, factored form is best, so we'll just copy over the other factors. And we can remove the common factor, as long as we add the qualifier that this a quality is true only for x not equal to 1. But since we're taking the limit, we don't care what happens at x equals 1, so we can replace the expression in the limit. And at x equals 1, the algebraic expression is defined, so the limit is the function value.