^{2}-3ax+x-a-1)/ x

^{2}-2x-3

Find the limit as "x" approaches 3

F(x)= (2x^{2}-3ax+x-a-1)/ x^{2}-2x-3

**I already know that if you substitute in 3 for x, the denominator will be 0 and therefore it cannot exist. I'm wondering if someone can find a way to factor it so that there can be a hole or removable discontinuity and therefore the limit may be able to exist.

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Here is another way to solve the problem.

Find a value of a that makes the numerator equal to 0 at x = 3. That sets up an indeterminate form at x = 3, which we can evaluate using L'Hopital's rule to find the limit at x = 3.

2x^{2} + (1 - 3a)x - (a + 1) = 0

Solve for a when x = 3, and get

20 = 10a

a = 2

The numerator becomes

2x^{2} - 5x - 3 = 0

Note that

Lim_{x→3 }[(2x^{2} - 5x - 3) / (x^{2} - 2x - 3)]

is indeterminate. Apply L'Hopital's rule and get

(4x - 5) / (2x - 2) → 7/4 as x → 3

Same result as in Victoria's solution, different method.

Hi R.

When you are finding a limit where the denominator turns out to be 0, then you have to find some other method than just plugging in the number x is approaching.

Factoring both the top and the bottom look like it might work here. Hopefully the (x-3) factor will cancel and the denominator will no longer be 0, and the limit will exist.

Start by factoring an "x" out of the middle two terms in the numerator, and factoring out a (-1) from the last two terms. Meaning:

2x^{2} - 3ax + x - a - 1 could be rewritten as:

The denominator factors into (x - 3)(x + 1).

So if we could factor the numerator so that it had a factor of (x-3), then the (x-3) on top would cancel with the (x - 3) on the bottom, and the limit WOULD exist.

So we are going to try to get

2x^{2} + (1 - 3a)x - (a + 1) = (2x + ???? ) (x - 3)

If we could, then we would have

lim_{x-->3}[(2x+???)(x-3)]/[(x-3)(x+1)]

The (x-3)s would cancel and the limit would be whatever is left when you plug in x=3

So lets replace the ???? with the variable N. Now we have

(2x + N)(x-3)

Foil this and get

2x^{2} - 6x + Nx - 3N

Gather like terms and get

2x^{2} + (N-6)x - 3N

Now go back to that big, bolded expression near the top, and if we set the coefficients of the x-terms equal. And set the constant terms equal, we should be able to find an "a" that works.

So

N - 6 = 1 - 3a and

3N = a + 1

This is now a system of equations with two unknowns, so I will put all of the variables on the left and the constants on the right and get

N + 3a = 7

3N - a = 1

Solve this and find that N = 1 and a = 2

That makes the big bolded numerator above

2x^{2} + (-5)x - 3 or 2x^{2} - 5x - 3

This factors into (2x + 1)(x - 3)

Now, finally the limit:

lim_{x-->3}[(2x+1)(x-3)]/[(x+1)(x-3)]

= lim_{x-->3}[(2x+1)]/[(x+1)] (the (x-3)'s cancel)

And that is the answer!

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## Comments

^{2}- 3ax + x - a - 1