FIND THE LIMIT!!! lim x--->3

sqrt(2x+3)- sqrt (3x)/(x^2-3x)
I keep getting zero is this right?

(√(2x+3)-√(3x))/(x^2-3x)

If you have l'Hospital's Rule as a tool, then you know that the limit is the same as

(1/√(2x+3) - 3/(2√(3x)) / (2x-3)
= (1/3 - 3/6)/3
= -1/18

To find the limit as x approaches 3 for the given expression, you can use algebraic simplification and factorization.

First, let's simplify the expression:
sqrt(2x + 3) - sqrt(3x) / (x^2 - 3x)

To combine the two square root terms, we need to rationalize the denominator. Multiply both the numerator and denominator by the conjugate of the denominator, which is (x + 3):

[(sqrt(2x + 3) - sqrt(3x))(x + 3)] / [(x^2 - 3x)(x + 3)]

Now we can expand the numerator:
(sqrt(2x + 3) - sqrt(3x))(x + 3) = (sqrt(2x + 3)(x + 3)) - (sqrt(3x)(x + 3))

Using the distributive property, expand both terms:
(sqrt(2x + 3)(x) + sqrt(2x + 3)(3)) - (sqrt(3x)(x) + sqrt(3x)(3))

Evaluate the expanded expression:
(x * sqrt(2x + 3) + 3 * sqrt(2x + 3)) - (x * sqrt(3x) + 3 * sqrt(3x))

Now we can simplify the expression further:
x * (sqrt(2x + 3) - sqrt(3x)) + 3 * (sqrt(2x + 3) - sqrt(3x))

Now let's factor out (sqrt(2x + 3) - sqrt(3x)):
(sqrt(2x + 3) - sqrt(3x))(x + 3)

Now we can cancel out the common factor from the numerator and denominator:
[(sqrt(2x + 3) - sqrt(3x))(x + 3)] / [(x^2 - 3x)(x + 3)]

We can now see that the (x + 3) terms in the numerator and the denominator will cancel each other.

Taking the limit as x approaches 3:
lim(x->3) [(sqrt(2x + 3) - sqrt(3x))(x + 3)] / [(x^2 - 3x)(x + 3)]

Now, if we substitute x = 3 into the simplified expression, we get:
[(sqrt(2(3) + 3) - sqrt(3(3)))(3 + 3)] / [(3^2 - 3(3))(3 + 3)]

This simplifies to:
[(sqrt(9) - sqrt(9))(6)] / [(9 - 9)(6)]

Since sqrt(9) = 3, the expression further simplifies to:
[(3 - 3)(6)] / [(0)(6)]

And simplifying this further, we get:
[0(6)] / [0]

At this point, we have an indeterminate form of 0/0.

To find the actual value of the limit, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form of 0/0 (or ∞/∞), we can take the derivative of the numerator and the denominator separately and then evaluate the limit again.

Taking the derivative of the numerator:
d/dx (sqrt(2x + 3) - sqrt(3x)) = (1/2)(2x + 3)^(-1/2) * 2 - (1/2)(3x)^(-1/2) * 3

Simplifying the derivative of the numerator:
(1/sqrt(2x + 3)) - (3/sqrt(3x))

Now taking the derivative of the denominator:
d/dx [(x^2 - 3x)(x + 3)] = (2x - 3)(x + 3) + (x^2 - 3x)

Simplifying the derivative of the denominator:
2x^2 + 3x - 9 + x^2 - 3x = 3x^2 - 9

Next, we take the limit again using the derivatives:
lim(x->3) [(1/sqrt(2x + 3)) - (3/sqrt(3x))] / [3x^2 - 9]

Substituting x = 3 into the expression:
[(1/sqrt(2(3) + 3)) - (3/sqrt(3(3)))] / [3(3)^2 - 9]

Simplifying it further:
[(1/sqrt(6 + 3)) - (3/sqrt(9))] / [3(9) - 9]

[(1/sqrt(9)) - (3/3)] / [18 - 9]

[1/3 - 1] / 9

(1/3 - 1) / 9 = (-2/3) / 9

Finally, we get the result:
-2/27