Hey...do check the wikipedia page ...http://en.wikipedia.org/wiki/L'Hôpital's_rule.....do get back if there is any problem you r facing
–
Tanya Gupta
·
2 years, 10 months ago

Log in to reply

L hospital's rule is used to solve indeterminate forms of the type 0/0 ,infinity/infinity type .Put simply,if the two functions f(x) &g(x) is defined in an interval containing' a ',except possibility at' a' such that lim f(x) =0 &lim g(x) =0
x turns to a x turns to a
but the derivatives f'(x)&g'(x) has finite limit at x=a
then the theorem state that lim f(x)/g(x) = lim f'(x)/g'(x) x turns to a x turns to a

for eg ;evaluate lim x^2 -4/x-2 ( form 0/0) x turns to 2
let f(x)=x^2 -4 , f'(x)= 2x
g(x) = x-2 g'(x) =1 hence lim x^2 -4/x-2= 2x/1 = 4 x turns to 2
for better reference , clarifications and proof u can refer CALCULUS by Anton ,Bivens and Stephen Davis
–
Santoshi Ganneda
·
2 years, 12 months ago

Log in to reply

L hospital's rule is used to solve indeterminate forms of the type 0/0 ,infinity/infinity type .Put simply,if the two functions f(x) &g(x) is defined in an interval containing' a ',except possibility at' a' such that
lim f(x) =0 &lim g(x) =0
x turns to a x turns to a
but the derivatives f'(x)&g'(x) has finite limit at x=a
then the theorem state that
lim f(x)/g(x) = lim f'(x)/g'(x)
x turns to a x turns to a

for eg ;evaluate lim x^2 -4/x-2 ( form 0/0)
x turns to 2
let f(x)=x^2 -4 , f'(x)= 2x
g(x) = x-2 g'(x) =1
hence lim x^2 -4/x-2= 2x/1 = 4
x turns to 2
for better reference , clarifications and proof u can refer CALCULUS by Anton ,Bivens and Stephen Davis
–
Adithya Maroli
·
2 years, 12 months ago

Log in to reply

Do I need to ? Ain't you know that ? ;)
–
Sudhanshu Pandey
·
2 years, 12 months ago

## Comments

Sort by:

TopNewestHey...do check the wikipedia page ...http://en.wikipedia.org/wiki/L'Hôpital's_rule.....do get back if there is any problem you r facing – Tanya Gupta · 2 years, 10 months ago

Log in to reply

L hospital's rule is used to solve indeterminate forms of the type 0/0 ,infinity/infinity type .Put simply,if the two functions f(x) &g(x) is defined in an interval containing' a ',except possibility at' a' such that lim f(x) =0 &lim g(x) =0 x turns to a x turns to a but the derivatives f'(x)&g'(x) has finite limit at x=a then the theorem state that lim f(x)/g(x) = lim f'(x)/g'(x) x turns to a x turns to a

for eg ;evaluate lim x^2 -4/x-2 ( form 0/0) x turns to 2 let f(x)=x^2 -4 , f'(x)= 2x g(x) = x-2 g'(x) =1 hence lim x^2 -4/x-2= 2x/1 = 4 x turns to 2 for better reference , clarifications and proof u can refer CALCULUS by Anton ,Bivens and Stephen Davis – Santoshi Ganneda · 2 years, 12 months ago

Log in to reply

L hospital's rule is used to solve indeterminate forms of the type 0/0 ,infinity/infinity type .Put simply,if the two functions f(x) &g(x) is defined in an interval containing' a ',except possibility at' a' such that lim f(x) =0 &lim g(x) =0

x turns to a x turns to a

but the derivatives f'(x)&g'(x) has finite limit at x=a

then the theorem state that lim f(x)/g(x) = lim f'(x)/g'(x) x turns to a x turns to a

for eg ;evaluate lim x^2 -4/x-2 ( form 0/0) x turns to 2

let f(x)=x^2 -4 , f'(x)= 2x

g(x) = x-2 g'(x) =1 hence lim x^2 -4/x-2= 2x/1 = 4 x turns to 2

for better reference , clarifications and proof u can refer CALCULUS by Anton ,Bivens and Stephen Davis – Adithya Maroli · 2 years, 12 months ago

Log in to reply

Do I need to ? Ain't you know that ? ;) – Sudhanshu Pandey · 2 years, 12 months ago

Log in to reply