Solving limits (calculus)?

Question

Hi,
please help me understand how to solve this problem:

Let f(x)= (3x^2 - 9x - 12) / (x^2 + 2x -24)

Find:

lim->4 f(x)

Thank you!

Answer 1

it should be obvious that if you plug in x=4, it is undefined. im not sure what youve learned so far, but theres 2 ways to go about this:
1)factor the top and bottom. in theory, something should be able to cancel out. then plug in the 4.

2)use la hopitals rule (im not sure if thats how its spelled. we usually call it 'la hospitals' rule. haha). find the deriviate of the top, then the derivative of the bottom. if it is still undefined, find the derivative again. (do the top and bottom as separate funtions. do NOT use the quotient rule!). then just plug in the 4, and youve got the limit!

EDIT:
heyy! thats not fair! the guy above me just wrote "factor the top, then factor the bottom" then he went back and solved it! ugh. anyway, i guess since he factored, ill use la 'hospitals' rule:

deriv of the top: 6x-9
deriv of the bottome: 2x+2

plug in 4: (6*4-9)/(2*4+2)
=15/10=3/2

Answer 2

First factor the top and the bottom.

(3x^2 - 9x - 12) = 3(x - 4)(x+1)
(x^2 + 2x -24) = (x - 4)(x+6)

So (3x^2 - 9x - 12) / (x^2 + 2x -24) =
3(x+1)/(x+6)

Then to get the limit as x->4, just plug in 4 to get:

3*5/10 = 3/2

Edit: Sorry, I did not know that was not allowed. I once had someone just put "e" and then after I had written a complete answer he wrote an answer just like mine (and by the way the answer wasn't e). I can attest to the fact that Laura had everything before the edit done before I had my answer, though I think I was first with the 3/2.

Answer 3

first you factor.

(3x^2-9x-12)/(x^2+2x-24)


Lets first factor the Numerator...factor a 3 out...

3*(x^2-3x-4) this factors into

3*(x+1)(x-4) now the Denominator


(x^2+2x-24)=(x+6)(x-4) putting together...


[3*(x+1)(x-4)]/[(x+6)(x-4)] both x-4 cancels left with

[3*(x+1)]/[(x+6)]

f(4)= 3*(4+1)/(4+6)=15/10=5*3/5*2=3/2.

now taking the limit as x->4 f(x)->3/2

Answer 4

If you directly subsitute 4, you get infinity/infinity which is an indeterminate form.
so you take 3 as a common factor in the numberator to get x^2 - 3x -4

now you have
f(x) = 3(x^2-3x-4)/(x^2+2x-24)

or f(x) ={ 3(x-4)(x+1) }/ {(x+6)(x-4)}
You get this by factorizing hte polynomial
then you cancell off (x-4) in numerator and denominator
then you get
3(x+1)/(x+6)

put the limit 4
you get the answer as 15/10 = 3/2

Answer 5

3x² - 9x - 12 = 3(x² - 3x - 4)
= 3(x + 1)(x - 4)

x² + 2x - 24 = (x + 6)(x - 4)

lim->4 f(x) = lim->4[3(x + 1) / (x + 6)]
= 3(5) / 10
= 3/2.

Answer 6

Given,
f(x) = (3x^2-9x-12)/(x^2+2x-24)
on writing (-9x) as ( - 12x + 3x) and (2x) as (+6x - 4x), we get,

f(x) = [(3x^2 - 12x + 3x - 12)/ (x^2 + 6x - 4x - 24)]

= [{3x (x - 4) + 3 (x - 4)}/ {x (x + 6) -4 (x + 6)}

= {(3x + 3)(x - 4)/(x + 6)(x - 4)}
The factor (x-4) is canceled and we get

f(x) = (3x+3)/(x + 6)
lim x>4 f(x) = (3*4 + 3)/(4 + 6)
= (15)/(10)

= 3/2

Answer 7

Factor the top and bottom of f(x), then you get,
f(x) = 3(x+1)(x-4) / (x-4)(x+6)
= 3(x+1)/(x+6)
lim x->4 f(x) = 3(4+1)/(4+6) = 15/10 = 3/2
Alternatively by L'Hopital's rule differentiate the top and bottom,
lim x->4 f(x) = lim x->4 [(6x-9)/(2x+2)] = 15 / 10 = 3/2

Hope this helps!

Answer 8

First, factor
3(x-4)(x+1)/(x+6)(x-4) ... (You need to get rid of the x-4 to remove infinities, or where the function is undefined)

= 3(x+1)/(x+6)

= 3x+3/x+6 = (3(4)+3)/(4+6) = 15/10 = 1.5

Does that sound good?

Answer 9

1) lim e^Ln(x) = Lim(x→6) x = 6 x→6 e^Ln(x) = x by definition of the Naperian Logarithms 2) lim g(x) Am I suppose to guess what g(x) is? x→3^+ 3) Lim |7x|/x = Lim(x→0+) 7|x|/x = 7 x → 0^+ Since x is not 0 nor negative then |x|/x = 1