find the limits algebraically?

Question

I'm having a problem with this seemingly easy limit question:

find the limit of (x^2-3x+2)/(x^3-4x) as x approaches 2 from the right, as x approaches -2 from the right, as x approaches 0 from the left, and as x approaches 1 from the right

easy to do it graphically but how do I do it algebraically?

the problem here is that I did all you guys are showing, and I get 1/8 for the limit as x -->2, but then if actually I graph the function on a calculator the limit is actually 1/4 as x approaches 2

omg, I see it, I was to lazy to check that the function I plugged in my calculator was the wrong one! thanks a lot you guys

Answer 1

Numerator = (x-2)(x-1)
Den = x*(x+2)(x-2)

Num / Den = (x-1) / [x(x+2)]
As x --> -2+,
limit = -3 / [-2 * (+0)]
= +∞

As x --> 1, the limit is 0 from either side.

Answer 2

The simplest way to find the limit for any equation when you are given a value for x is by substitution. Simply substitute the value into the equation and see what x equals.

However, as I'm sure you noticed, doing this for x = 2, -2 and 0 gives you an error. This is because all three of these values make the denominator 0, as you cannot divide by 0, making the fuction undefinded.

You can, thankfully, substitute in x = 1 though. Doing so gets you a value of 0. So the limit as x ~> 1 = 0.

For the others, factor the expression and cancel out all that you can: (x^2 - 3x + 2) / (x^3 - 4x) = [(x-2)(x-1)] / [x(x-2)(x+2)]

As you can see, the x-2 will cancel out, you get (x-1) / [x(x+2)]

This allows you to directly substitute in x = 2, giving you an answer of 1/8. So the Limit as x ~> 2 is 1/8.

The last two are a bit harder to get since direct substitution still will not work. This is where the phrases "from the right" and "from the left" become important.

For x ~> -2 from the right, pick a decimal VERY close to -2 that is on the right side of -2 on a number line, such as -1.9. Subtituting that in, you find that the fuction = 15.26.

Then pick another value even closer to -2 on the right side such as -1.99. You find the fuction = 150.25. With each value closer to -2, you are getting a larger and larger answer. So, the limit as x ~> -2 from the right is positive infinity.

You will follow a similar line of thinking for the limit as x ~> 0 from the left.

Answer 3

(x²-3x+2)/(x³-4x)

First, factorize the rational function:

(x-1)(x-2)/(x(x-2)(x+2))

Which for x≠2 is:

(x-1)/(x(x+2))

So we have:

[x→2⁺]lim (x²-3x+2)/(x³-4x) = [x→2⁺]lim (x-1)/(x(x+2)) = 1/(2*4) = 1/8

[x→-2⁺]lim (x²-3x+2)/(x³-4x) = [x→-2⁺]lim (x-1)/(x(x+2)) = ∞ (since the denominator approaches 0 and the numerator approaches a positive number, the absolute value approaches ∞, and since the function is positive just to the right of -2, the function approaches ∞ as well)

[x→0⁻]lim (x²-3x+2)/(x³-4x) = [x→0⁻]lim (x-1)/(x(x+2)) = ∞ (absolute value approaches ∞, function is positive just to the left of 0)

[x→1⁺]lim (x²-3x+2)/(x³-4x) = [x→1⁺]lim (x-1)/(x(x+2)) = 0/(1*3) = 0

And we are done.

Edit: Well, you must be doing something wrong with your graphing calculator then, because the limit is 1/8. Take a look at this graph centered at x=2: http://img299.imageshack.us/img299/2319/limitlc4.png . Note that near x=2 the y-value is near .125 = 1/8.

And who's the jerk that gave me a thumbs down?

Answer 4

Factor (x^2-3x+2)/(x^3-4x)
= (x-2)(x-1)/(x)(x-2)(x+2)
For lim (x-2)'s cancel out, so you get
(x-1)/(x)(x+2)
2 isnt a root so plug in 2
1/(2)(4)
so the lim is 1/8

Answer 5

factor the numerator and denominator so that

(x^2 - 3x + 2)/(x^3 - 4x) = (x - 2)(x - 1)/[x(x^2 - 4)] =
(x - 2)(x - 1)/[x(x - 2)(x + 2)] = (x - 1)/[x(x+2)]

Now taking each limit should be straight forward for you.

Hope this helps!