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# Is there a number to "a" such that the limit exists? If so, find the value of "a" and find the limit. If not, explain why.

Find the limit as "x" approaches 3
F(x)= (2x2-3ax+x-a-1)/ x2-2x-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.

Please check the signs on the "a" and the "1."
Is the numerator

2x2 - 3ax + x - a - 1

correctly stated?  Having 2 terms with x looks strange.  Please check the original problem.

### 2 Answers by Expert Tutors

Arturo O. | Experienced Physics Teacher for Physics TutoringExperienced Physics Teacher for Physics ...
5.0 5.0 (66 lesson ratings) (66)
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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.

2x2 + (1 - 3a)x - (a + 1) = 0

Solve for a when x = 3, and get

20 = 10a

a = 2

The numerator becomes

2x2 - 5x - 3 = 0

Note that

Limx→3 [(2x2 - 5x - 3) / (x2 - 2x - 3)]

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

(4x - 5) / (2x - 2) → 7/4 as x → 3

a = 2
The limit is 7/4.

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

Victoria V. | Math Teacher: 17+ Years teaching/tutoring Calc, Algebra 2, Trig, GeomMath Teacher: 17+ Years teaching/tutorin...
4.9 4.9 (35 lesson ratings) (35)
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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:

2x2 - 3ax + x - a - 1   could be rewritten as:

## 2x2 + (1-3a)x  - (a + 1)

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

2x2 + (1 - 3a)x - (a + 1)  = (2x +  ???? ) (x - 3)

If we could, then we would have

limx-->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

2x2 - 6x + Nx - 3N

Gather like terms and get

2x2 + (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

2x2 + (-5)x - 3   or    2x2 - 5x - 3

This factors into  (2x + 1)(x - 3)

Now, finally the limit:

limx-->3[(2x+1)(x-3)]/[(x+1)(x-3)]

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