I want to work on a quick math program but i can't remember the formula for getting the derivative of a fraction such as
Lim X > 0 X^4/ X1
Any help from the calculas kings appreciated.
:alright:
Lim X > 0 X^4/ X1
Any help from the calculas kings appreciated.
:alright:
As I remember it, for any fraction:
IE, the derivative of the numerator, divided by the derivative of the denominator, evaluated at the point of interest.
F(x) = N(x) / D(x)
May be evaluated by:
d N(x) d D(x)
F(Xo) =  / 
dx  dx 
Xo Xo
IE, the derivative of the numerator, divided by the derivative of the denominator, evaluated at the point of interest.
To check your results:
Hi !
In general:
If you have two functions, g(x) and h(x), and g'(x) is the deriviation of g by x and h'(x) of h(x) by x then the basic rules are:
f(x) = g(x) + h(x) => f'(x) = g'(x) + h'(x)
f(x) = g(x) * h(x) => f'(x) = g'(x) * h(x) + g(x) * h'(x)
f(x) = g(x) / h(x) => f'(x) = (g'(x) * h(x)  g(x) * h'(x)) / (h(x)^2)
f(x) = g(h(x)) => f'(x) = (dg / dh) * h'(x)
Greetings, CALEB
In general:
If you have two functions, g(x) and h(x), and g'(x) is the deriviation of g by x and h'(x) of h(x) by x then the basic rules are:
f(x) = g(x) + h(x) => f'(x) = g'(x) + h'(x)
f(x) = g(x) * h(x) => f'(x) = g'(x) * h(x) + g(x) * h'(x)
f(x) = g(x) / h(x) => f'(x) = (g'(x) * h(x)  g(x) * h'(x)) / (h(x)^2)
f(x) = g(h(x)) => f'(x) = (dg / dh) * h'(x)
Greetings, CALEB
hi all,
Limit is not like derivative, The derivative is a special case of the limit.
The derivative is a limit like this:
lim h>0 (F(X+H)F(X))/h
But limits in general differs.
There are some laws for it:
1 by using direct substitution like in your case:
lim x>0 X^4/X1=0^4/01=0/1=0
Therefor the limit=0
2If the limit gives 0/0 by direct substitutions there will be other ways to deal with the limit
Limit is not like derivative, The derivative is a special case of the limit.
The derivative is a limit like this:
lim h>0 (F(X+H)F(X))/h
But limits in general differs.
There are some laws for it:
1 by using direct substitution like in your case:
lim x>0 X^4/X1=0^4/01=0/1=0
Therefor the limit=0
2If the limit gives 0/0 by direct substitutions there will be other ways to deal with the limit
eet_1024: you do you create this equations? is there a program i can create such equations easily?
nop
nop
it equals to 0 becuase when x > 0 . image x = 0.0000000001
and when you will do x^4 the number will get lower and lower .
in your equation the numerator equals to zero.
but if your x> 1 . you were in a problem becuase the denominator were equals to zero( and this is another story)
check out
nice math program
math editor
p.s
i'm sorry for my english and hope you understood my explantion
and when you will do x^4 the number will get lower and lower .
in your equation the numerator equals to zero.
but if your x> 1 . you were in a problem becuase the denominator were equals to zero( and this is another story)
check out
nice math program
math editor
p.s
i'm sorry for my english and hope you understood my explantion
Maybe I should have given the complete answer.
For limit problems such as this, it is a legitimate technique to see 'how fast' the numerator and denominator get to the point. This is why the lim x>0 sin(x)/x =1 (cos(0)/1 = 1/1 = 1)
Anyway,
d
 x^4 = 4x^3
dx
d
 x1 = 1
dx
thus 4x^3/1 = 0/1 = 0

x=0
For the limit as x>1, this would evaluate to 4.
For limit problems such as this, it is a legitimate technique to see 'how fast' the numerator and denominator get to the point. This is why the lim x>0 sin(x)/x =1 (cos(0)/1 = 1/1 = 1)
Anyway,
d
 x^4 = 4x^3
dx
d
 x1 = 1
dx
thus 4x^3/1 = 0/1 = 0

x=0
For the limit as x>1, this would evaluate to 4.
Hi Ernie !
Your way is the rule by l'Hospital but you can use it only if the fraction of the two functions is undefined (either 0/0 or oo / oo )
So you can use it for
lim x>1 (x^4/(x1))
but not to get
lim x>0 (x^4/(x1))
The rule itself says if you have lim x>x0 ( f(x) / g(x) ) and its form is ( 0/0 ) or ( 00 / 00 ) then
lim x>x0 ( f(x) / g(x) ) = lim x>x0 ( f'(x) / g'(x) )
if it exists !
Greetings, CALEB
Your way is the rule by l'Hospital but you can use it only if the fraction of the two functions is undefined (either 0/0 or oo / oo )
So you can use it for
lim x>1 (x^4/(x1))
but not to get
lim x>0 (x^4/(x1))
The rule itself says if you have lim x>x0 ( f(x) / g(x) ) and its form is ( 0/0 ) or ( 00 / 00 ) then
lim x>x0 ( f(x) / g(x) ) = lim x>x0 ( f'(x) / g'(x) )
if it exists !
Greetings, CALEB
You can use l'Hopital wherever you want. It helps to get you out of situations where the limit is undefined, but its not limited to those situations.
But in this case theres no need, don't forget the rule of thumb; if the limit makes sense, use it.
So Lim X > 0 X^4/ X1 becomes 0^4 / 01 = 0 / 1 = 0
But in this case theres no need, don't forget the rule of thumb; if the limit makes sense, use it.
So Lim X > 0 X^4/ X1 becomes 0^4 / 01 = 0 / 1 = 0
Hi E?in !
You get wrong ! You cannot use l'Hospital everywhere you want to get the right solution ! It is only for those undefined limits. But you can transform terms like oo  oo or 0^0 to a fraction of functions and then use l'Hospital !
Greetings, CALEB
You get wrong ! You cannot use l'Hospital everywhere you want to get the right solution ! It is only for those undefined limits. But you can transform terms like oo  oo or 0^0 to a fraction of functions and then use l'Hospital !
Greetings, CALEB
nop:
Plain and cheap paint.exe :)
Plain and cheap paint.exe :)