What Is the Correct Limit Calculation for limx->3 (x^2f(x) - 18) / (x-3)?

[Zargawee]

[SOLVED] Small Limits Question

Hi There,
I have this Simple Question In Limits :

If lim_x->3 (f(x) - 2) / (x - 3) = 7
Then lim_x->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

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

But I also solved in this way :

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


What's the wrong with the first One ?
Please help.

 

[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 ?
Please help.

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

 

[Soroban]

Hello, Zargawee!

I found an answer. (Please check my work!)

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

We're asked to find: lim_x->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:
lim_x->3(x^2) * lim_x->3 [f(x) - 2]/(x - 3) + lim_x->3 2(x + 3)

We are given that the middle limit is 7.

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

 

[Hurkyl]

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

lim_x→3 f(x) = 2


Soroban's answer looks right.

 

[StephenPrivitera]

<--- 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.