Ask Me Help Desk - Calculus . About limits

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Calculus . About limits

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 you..

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

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!

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

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 a lot..

Hi ebaines! I wanted to know something...

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

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.

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

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. $

uha..

ebaines.., Thanks..

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

our teacher didn't give it to us..

THANKS AGAIN!.

=)..

by the way.. there is a mistake..

the numerator will be 41/3

and the denominator = 21

SO

the final ANS..

41/3 / 21

= 41/63

Q: when we can USE this rule ?

There is another way, which does not require using l'Hospital's rule. If you haven't gotten that far in your math course then I suspect that this is what your teacher is looking for you to do:

Note that the denominator can be simplified as 7*(3x+2). Which makes me wonder - can the numerator be factored into a quadratic times (3x+2)? If so, then you can eliminate the (3x+2) from both numerator and denominator, and then this problem becomes trivial. Try it, and you'll see that it can indeed. You get the same answer as before, but without using l'Hospital.

Quote:

Originally Posted by malakii211

by the way .. there is a mistake ..

the numerator will be 41/3

and the denominator = 21

SO

the final ANS ..

41/3 / 21

= 41/63

Right!

Quote:

Originally Posted by malakii211

Q: when we can USE this rule ?

Whenever you want - it comes in most handy when both the numerator and denominator are tending towards 0, or both are tending towards infinity. Also, if after applying l'Hospital you find that both numerator and denominator are still heading to 0 (or infinity), you can apply the rule again, and again, and again - as many times as you need!

Quote:

Originally Posted by ebaines

There is another way, which does not require using l'Hospital's rule. If you haven't gotten that far in your math course then I suspect that this is what your teacher is looking for you to do:

Note that the denominator can be simplified as 7*(3x+2). Which makes me wonder - can the numerator be factored into a quadratic times (3x+2)? If so, then you can eliminate the (3x+2) from both numerator and denominator, and then this problem becomes trivial. Try it, and you'll see that it can indeed. You get the same answer as before, but without using l'Hospital.

I've tried that, no use. I think we are bound to use l'Hospital's rule.

Quote:

Originally Posted by Unknown008

I've tried that, no use. I think we are bound to use l'Hospital's rule.

Now you're going to make me back up what I wrote! OK, here goes:
$ 6x^3-8x^2-5x+2 = (2x^2-4x+1)(3x+2) $

So:

$ \lim _{x \to {\frac {-2} 3} } ( \frac {6x^3-8x^2-5x+2} {21x+14} ) = \lim _{x \to {\frac {-2} 3}} \frac {(2x^2-4x+1)(3x+2)} {7(3x+2)} = \lim _{x \to {\frac {-2} 3}} \frac {2x^2 - 4x +1} 7 = \frac {2*(\frac {-2} 3 ) ^2 + 4(\frac {-2} 3) +1} 7 = \frac {41} {63} $

uha..

Yup it's correct

our teacher gave us another way which is :

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

=

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

= 41/63..

There are many ways.. but I think in our college we connot use the role because it's not in the book.. but sure I'll use it for checking..

Quote:

Originally Posted by malakii211

uha ..

Yup it's correct

our teacher gave us another way which is :

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

=

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

= 41/63 ..

There are many ways .. but I think in our college we connot use the role because it's not in the book .. but sure I'll use it for checking ..

The rule.. :rolleyes: BY MISTAKE..

Thanks.. :rolleyes:.. for helping..

the last Q

How we can use the rule if we have an exercise like :

lim┬(x→0)⁡〖(1-√(1+x))/x〗

Quote:

Originally Posted by malakii211

Thanks.. :rolleyes: .. for helping ..

the last Q

How we can use the rule if we have an exercise like :

lim┬(x→0)⁡〖(1-√(1+x))/x〗

If I follow your notation correctly, you have:

$ \lim _{x \to 0} \frac { 1 - sqrt{1+x} } x $

Is that right?

Take the derivative of top and bottom, and apply the limit:

$ \lim _{x \to 0} \frac { - \frac 1 2(1+x)^{- \frac 1 2}} 1 = - \frac 1 2 $

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