|
Something easy, but I always second guess myself. For example:
lim x-> 2
3x^2 + 5
The answer is 17, just plug in, and that's the limit right? What if it is;
lim x-> -2
3x^2 + 5
Plug it in again and I get 17 again, right?
Okay, enough with second-guessing myself on those; how about something like this?
lim
x->5
(x^2-3x-10) / (x-5)
clearly it turns out to be 0/0, but not all 0/0 have limits, vice-versa. So let's do this:
( (x+2)(x-5) ) / (x-5)
(x+2) / 0
5+2 / 0
7 / 0
the answer says 7 is the limit but 7 / 0 is infinite, how can this be? Can someone explain? I also plugged the function into the calculator and I get a straight slant line.
Moving on into infinite limits, this confuses me, I mean, I understand it but it still confuses me;
lim x-> ∞
(3x^3+2) / (9x^3-2x^2+7)
This turns out to be 1/3 as it's 3/9 reduced, how? I know as numbers tend to go very high, the +7 & +2 are negligible but what about the 2x^2? Someone explain to me how it is 1/3 ?
Thanks, hope my questions were clear!
|
...what grade are you in?
|
( (x+2)(x-5) ) / (x-5)
= (x+2)/1
So its 7.
When you divide top and bottom by x-5 you get 1 not 0.
(3x^3+2) / (9x^3-2x^2+7)
divide top and bottom by x^3, then you get.
(3+2/x^3) / (9-2/x+7/x^3)
and taking the limit, you get 3/9 =1/3
Hope that helps.
|
lim x-> ∞
(3x^3+2) / (9x^3-2x^2+7)
divide everything by x^3
lim x-> ∞
(3+2/x^3) / (9-2/x+7/x^3)
as x-> ∞, 3/9 = 1/3
|
Oops, stupid me, sorry on;
( (x+2)(x-5) ) / (x-5)
= (x+2)/1
So its 7.
For this;
lim x-> ∞
(3x^3+2) / (9x^3-2x^2+7)
(3+2/x^3) / (9-2/x+7/x^3)
How come the 2/x^3 and the 2/x & 7/x^3 don't count? That's what I don't get, sorry for not being so clear on that.
|
Canada7170 Posts
|
Calgary25986 Posts
|
lim x-> ∞
(3x^3+2) / (9x^3-2x^2+7)
(3+2/x^3) / (9-2/x+7/x^3)
then 2/x^3 goes to 0, 2/x goes to 0, 7/x^3 goes to zero so all thats left is 3/9=1/3
|
United States24723 Posts
Don't feel bad I made stupider mistakes in my dead bodies blog.
|
Zortch, do you graphically know that or? Because 2/x^3 is an asymptote that goes to 0, or you plug something in, etc?
|
okok um thing about 2/98321648217354018723087236598417651874650981465098164325164387561437 that...a really really really big number. 2 over that is practically zero.
So for 2/x as x goes to infinity, 2/x goes to zero because x is so huge.
Graphically if you graph 2/x and check out what happens as x gets huge, the function will be going to zero.
|
A limit is simply the value to which a function f(x) converges when x->a
Your first example lim x-> 2 (3x^2 + 5) is on a function that is continuous everywhere so naturally the limit will be the same as if you just "plugged in" x=a.
The kind of limits you call 0/0 or in other symbols 0*∞ are different though. One factor grows towards infinity and the other goes infinitly close to 0, so what is the product? It depends on how fast they grow in relation to each other, and that is what you can calculate.
It might be helpful to plot your functions and look at the horizontal line where the limit is.
|
On May 22 2008 06:49 Zortch wrote:
okok um thing about 2/98321648217354018723087236598417651874650981465098164325164387561437 that...a really really really big number. 2 over that is practically zero.
So for 2/x as x goes to infinity, 2/x goes to zero because x is so huge.
Graphically if you graph 2/x and check out what happens as x gets huge, the function will be going to zero.
Thanks for the clear explanation. Negative infinite would act the same way right? Like 2/-9999999.
|
Yea, negative infinity is the same, 2/-999999 or -2/9999999 are both very close to zero, just from the below the axis if you wanna thing graphically.
|
|
|
|
|
On May 22 2008 09:18 geometryb wrote:
use l'hopital's rule
|
|
|
|
|
|
EPT
