Calculus . about limits

malakii211 · #1 Oct 7, 2009, 09:39 AM

Hi every one ..

I have a problem with an exercise in ( limits )

I don't know what i should to do ..

lim( x-> -2/3 , (6x^3 - 8x^2 - 5x + 2) / (21x + 14))

!!

i need the answer quickly ..

c ya ..

malakii211 · Oct 7, 2009, 09:53 AM

I want to know ( understanding ) how to solve this exercise ..!!

Unknown008 · Oct 7, 2009, 10:23 AM

Hi malakii211! Welcome to AMHD!

I'm sure you'll get a response. I would like to help, but unfortunately, I've not studied limits like this at school yet. There are other members here though who will be able to answer you, as soon as they come online. Be a little more patient.

Thanks!

ebaines · Oct 7, 2009, 10:35 AM

You can use l'Hospital's rule to solve this:
$ \lim _ {x \to c} \ \frac {g(x)} {f(x)} = \lim _ {x \to c} \ \frac {g'(x)} {f'(x)} $

See: l'Hôpital's rule - Wikipedia, the free encyclopedia

malakii211 · Oct 7, 2009, 10:40 AM

Unknown008 :

however .. :P thank you aloooot .. I will don't worry .. but seriously i was afraid because i have to do it today ..

ebaines !!

amm Can You explain
please .. I mean to write the steps..

and thank you alot ..

Unknown008 · Oct 7, 2009, 10:42 AM

Hi ebaines! I wanted to know something...

If it's so, there's only to replace x by -2/3 in the 'equivalent derivative'?

ebaines · Oct 7, 2009, 10:49 AM

You would apply l'Hospital's rule like this:

For the numerator you have:
$ g(x) = 6x^3-8x^2-5x+2 $

So it's derivative is:
$ g'(x) = 18x^2 - 16x-5 $

For x = -2/3 you have
$ g'(2/3) = 16*(\frac {-2} 3)^2 - 16*(\frac {-2 }3) - 5 = 5 \frac 2 3 $

Now do the same process for the denominator, and see what you get for the derivative of the denominator evaluated at -2/3.

Then divide g'(-2/3) by f'(-2/3). Post back and tell us what you get.

Unknown008 · Oct 7, 2009, 10:52 AM

Typo ebaines... you typed 16 instead of 18 in the third LaTeX line.

ebaines · Oct 7, 2009, 10:55 AM

Quote:

Originally Posted by Unknown008

Hi ebaines! I wanted to know something...

If it's so, there's only to replace x by -2/3 in the 'equivalent derivative'?

Yep. For example, to find the limit as x approaches zero for sin(x)/x, you do it like this:

$ \lim _ {x \to 0} \frac {sin(x)} x = \lim _{x \to 0} \frac {\frac {dsin(x)} {dx}} {\frac {dx} {dx}} = \lim _{x \to 0} \frac {cos(x)} 1 = \frac {cos(0)} 1 = 1. $

malakii211 · Oct 7, 2009, 11:33 AM

uha ..

ebaines..,, Thanks ..

so EASY ! but unfortunately it's the first time that i had seen this rules ..

our teacher didn't give it 2 us ..

THANKS AGAIN !!...

=) ..



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