Hi studentThis is of the type f(x)g(x) which is of form 1 such that f(x)1 and g(x)Solution of such form is give by e g(x)(f(x)-1)When you will subsitute appropriate values of f(x) and g(x) you can get the answer.

Differential Calculus> Prove that...

Prove that

Shibashis Mallik , 11 Years ago

Grade 12

5 Answers

Harsh Patodia

Hi student

This is of the type $\lim_{x\rightarrow 0}$ f(x)^g(x) which is of form 1such that f(x) $\rightarrow$ 1 and g(x) $\rightarrow$ $\infty$

Solution of such form is give by e^g(x)(f(x)-1)

When you will subsitute appropriate values of f(x) and g(x) you can get the answer.

Last Activity: 11 Years ago

Shibashis Mallik

Is it necessary to subtract 1 from f(x) in the form e^g(x)(f(x)-1)?

Last Activity: 11 Years ago

Harsh Patodia

Hi Student,

Basically subtraction of 1 comes from the derivation. So it must be done.
If you want to know the derivation please reply on this thread otherwise its not necessary. You can remember the form the apply this directly.

Last Activity: 11 Years ago

Shibashis Mallik

Yes, why not? I definitely want to know the derivation.

Last Activity: 11 Years ago

Harsh Patodia

let y = f(x)^g(x)
Taking log on both sides
log y= g(x) log(f(x))
Taking antilog on both sides
y = e^{g(x) logf(x)}

Consider log f(x)= log ( 1 + (f(x)-1))
Expansion of log ( 1 + x)= x – x²/2 + x³/3 . . . .
log ( 1+ (f(x)-1)) = (f(x) – 1) – (f(x)-1)²/2 . . . . . .
since f(x) is tending to f(x) – 1 will be very small so higher are neglected
this means log ( 1+ (f(x)-1) = f(x) – 1 if f(x) tends to 1
The above expression becomes
y= e^g^{(x) ( f(x) -1)}