### Video Transcript

Find the limit of the function, the square root of π₯ minus two all over negative five π₯ squared plus 23π₯ minus 12 as π₯ approaches four.

The first thing we should consider is if we can evaluate this limit by using direct substitution. We know that we can evaluate the limit of a function π by direct substitution if the function π is polynomial which includes any constant function. If the function π is a power function, so π of π₯ is equal to π₯ to the power of π, where π is any real number. And if the function π is the sum, difference, product, or quotient of functions for which direct substitution works.

In our case, we see that the square root of π₯ is a power function where π equal to one-half. We can also see that two is a constant function and therefore weβll work with direct substitution. Since we know that we can use direct substitution on the square root of π₯ and the constant function too, we can take the difference of these two functions and still use direct substitution. So the numerator in our question will work with direct substitution.

Next, we see that the denominator of the function in our question is a polynomial. And so this will also work with direct substitution. Finally, since we have shown that we can use direct substitution on both the numerator and the denominator of the function in our question, we have shown that the function in our question is the question of functions for which direct substitution works. So we have justified that we can attempt to evaluate this intimate by using direct substitution.

Letβs clear some space. We see that our limit is when π₯ approaches four. So we substitute this value of four into our function. If we evaluate the numerator, we see that the positive square root of four is two. So we will have two minus two. And if we calculate the denominator, we will have negative 80 plus 92 minus 12. If we then calculate this, we will see that this gives us a zero divided by zero. So attempting to evaluate this limit by direct substitution has led us to an indeterminate form. This means weβre going to need to attempt to manipulate the expression into a form which we can evaluate the limit.

So letβs clear some space for this. We could attempt to factorize the quadratic in the denominator. In fact, we already found that by substituting the value π₯ equals four into a quadratic, this gave us a value of zero. So by the factor theorem, we must have that π₯ minus four is a factor. Hence, we can write out the limit in our question in following form, where ππ₯ plus π is equal to the unknown factor of the quadratic. We see that the coefficient of π₯ squared in our quadratic is negative five. This must be equal to the coefficient of π₯ squared that we get by multiplying our factors together. So we see that ππ₯ squared is equal to negative five π₯ squared. Hence, we must have that π is equal to negative five. Similarly, we must also have that negative 12 is equal to negative four multiplied by π. Therefore, π is equal to three.

This hasnβt changed anything yet. If we were to use direct substitution at this point, we would still get an indeterminate form. What we want at this point is to have a factor of π₯ minus four in the numerator which we could use the cancel the factor of π₯ minus four in the denominator. Fortunately, we can do this by multiplying by the conjugate of root π₯ minus two. We multiply both the numerator and the denominator by this conjugate so that the output of the function remains unchanged. We multiply out the denominator. And we can see that the numerator is multiplying the factors of a difference between squares. This gives us a numerator of the square root of π₯ squared minus two squared, which we can then simplify to just π₯ minus four.

Now, we know that if the functions π and π are equal at all points of an interval, except for the point π₯ is equal to π, and that the limit of π as π₯ approaches π is equal to πΏ. Then we can conclude that the limit of π as π₯ approaches π is also equal to πΏ. We can use this to justify canceling out our shared factor of π₯ minus four, since this will not change the output of the function at any point except for when π₯ is equal to four.

This gives us the following limit to evaluate. We see that the function in this limit is comprised of the product, quotient, and summation of polynomials and power functions. So we can evaluate this limit by direct substitution. We substitute the value of π₯ equals four into our function, giving us the following expression which if we then calculate, we will find to be negative one divided by 68.