# Small Limits Question

#### Zargawee

[SOLVED] Small Limits Question

Hi There,
I have this Simple Question In Limits :

If limx->3 (f(x) - 2) / (x - 3) = 7
Then limx->3 (x2f(x) - 18) / (x-3) = ??

I solved the question this way :
Since the denominator equals Zero , and limit exists , then the numerator equals zero .
[4]f(x) - 2 = 0 ---> f(x) = 2

limx->3 (x2f(x) - 18) / (x-3) =
limx->3 (2x2 - 18) / (x-3) =
limx->3 2(x2- 9) / (x-3) =
limx->3 2(x-3) (x+3) / (x-3) =
limx->3 2(x+3) =
2 (3+3) = 12

But I also solved in this way :

limx->3 (x2f(x) - 18) / (x-3) =
limx->3 (x2f(x) - 2x2 + 2x2 - 18) / (x-3)
limx->3 (x2f(x) + (2x2) /(x-3) - 2x2 - 18) / (x-3)
limx->3 (x2) (f(x) + 2) /(x-3) - 2(x2 - 9) / (x-3)
limx->3 (x2) (f(x) + 2) /(x-3) - 2(x2 - 9) / (x-3)
limx->3 (x2) 7 - limx->3 2(x-3)(x+3) / (x-3)
limx->3 (x2) 7 - limx->3 2(x-3)(x+3) / (x-3)
32 * 7 + limx->3 (x+3)
(9 * 7) + (3 + 3) = 63 + 12 = 75

What's the wrong with the first One ?

#### StephenPrivitera

Originally posted by Zargawee
Since the denominator equals Zero , and limit exists , then the numerator equals zero .
[4]f(x) - 2 = 0 ---> f(x) = 2

What's the wrong with the first One ?
I'm not sure about this reasoning. Your limit looks very similar to a difference quotient for derivatives. You could write limx-->a(f(x)-f(a))/(x-a)=f'(a). In this case, f'(3)=7, f(3)=2. You want to find limx-->ax2*f'(a)=a2f'(a)=9*7=63

#### Soroban

Hello, Zargawee!

We are given: limx->3 [f(x) - 2]/[x - 3] = 7

We're asked to find: limx->3[x2f(x) - 18]/[x - 3]

In the numerator, add 2x2 and subtract 2x2:
x2f(x) - 18 + 2x2 - 2x2 = x2[f(x) - 2] + 2(x2 - 9)

We have: x2[f(x) - 2]/(x - 3) + 2(x2 - 9)/(x - 3)

The second term reduces: 2(x - 3)(x + 3)/(x - 3) = 2(x + 3)

Then we have: x2[f(x) - 2)/(x - 3) + 2(x + 3)

Taking limits, we have:
limx->3(x^2) * limx->3 [f(x) - 2]/(x - 3) + limx->3 2(x + 3)

We are given that the middle limit is 7.

Therefore, the answer is: (32)(7) + 2(6) = 75

Last edited by a moderator:

#### Hurkyl

Staff Emeritus
Gold Member
Hi There,
I have this Simple Question In Limits :

If limx->3 (f(x) - 2) / (x - 3) = 7
Then limx->3 (x2f(x) - 18) / (x-3) = ??

I solved the question this way :
Since the denominator equals Zero , and limit exists , then the numerator equals zero .
f(x) - 2 = 0 ---> f(x) = 2
That's not quite right; what is true is that

limx&rarr;3 f(x) = 2

<--- ashamed

#### Zargawee

Thanks all,
But I think that I solved the question as the same as Soroban solved it (Look at my first post)

Thanks again.

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