Limit as x approaches -2: Why is the result 1/12 instead of 1?

[Jules18]

Limit is 0/0 ??

Homework Statement

limit as x approaches -2 of:

(x+2)/(x³+8)

The Attempt at a Solution

I plugged -2 into the eq'n and I got 0 for both the numerator and denomenator, so I thought the answer would be 1.


Apparently it's supposed to be 1/12 ...
is the text wrong?

 

[DMOC]

You need to factor (x^3 - 8).

I remember my PreCalc teacher telling me that if I get 0/0, I can always factor something out.

 

[Office_Shredder]

\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{ \lim_{x \rightarrow a} f(x)}{ \lim_{x \rightarrow a} g(x)}

holds only if \lim_{x \rightarrow a} g(x) is not equal to zero. So you can't just split the limit up and get 0/0 (which isn't defined to be 1 anyway).

If x³+8 is equal to 0 at x=-2, it means you can factor an (x-(-2)) = x+2 from it (basic fact about polynomials)

 

[popo902]

did you learn l'hopitals rule?
if you get 0/0 or infinty or undefined, can't you just get the derivative of the top and bottom
so you got 0/0
then f(x)/g(x)
find f'(x)/ g'(x) and keep getting the derivative of top and bottom until you get an actual value when you plug in a, or in this case, -2
am i wrong or...what?

 

[hanelliot]

L'hopital rule is fine of course.. you different top/bottom separately and get 1/3x^2, which is 1/12 if you plug in -2.

 

[DrMath]

if you get "0/0" - you should always use L'Hopital Rule.
Note that "0/1" or "1/0" results you shouldn't use... :)

 

[Mark44]

Homework Statement

limit as x approaches -2 of:

(x+2)/(x³+8)

The Attempt at a Solution

I plugged -2 into the eq'n and I got 0 for both the numerator and denomenator, so I thought the answer would be 1.


Apparently it's supposed to be 1/12 ...
is the text wrong?

L'Hopital's Rule is overkill for this problem, and is therefore not needed. The denominator can be factored into (x + 2) times a quadratic. The sum or difference of two cubes can be factored as follows:

a³ + b³ = (a + b)(a² - ab + b²)
a³ - b³ = (a - b)(a² + ab + b²)

 

[DMOC]

^
Agreed.

 

[HallsofIvy]

In fact, any time a polynomial, P(x), say, equals 0 for x= a, it must have x- a as a factor. You can use "long division" or "synthetic division" to determine the other factor.

 

[Dale]

L'Hopital's Rule is overkill for this problem, and is therefore not needed.

I don't know about that. L'Hopital's rule is more general (applies to non-polynomials) and differentiation is easier than factoring. Given the choice between an easy general solution and a difficult specialized solution I will pick the easy general approach every time.

 

[Mark44]

I don't know about that. L'Hopital's rule is more general (applies to non-polynomials) and differentiation is easier than factoring. Given the choice between an easy general solution and a difficult specialized solution I will pick the easy general approach every time.

Notice that I said "for this problem." Also, factoring is almost always taught before differentiation, so if the OP hasn't been exposed to differentiation yet, then L'Hopital's Rule would be out of the question, regardless of its generality.