simple

thanx a lot for the proof

And about why Lt x→0 f(x)^g(x) ≡ e^{Lt x→0 [f(x)-1].g(x)} when f(x) →1 and g(x) →∞ as x→0

First, we must have that the limit (f(x) - 1)g(x) as x→0 is finite

Let y = f(x)^g(x). Hence log y = g(x) log f(x) = [log (1+f(x) - 1)/f(x) - 1] [g(x) (f(x)-1)]

Hence Lt _x→0 log y = log Lt _x→0 y (since log is a continuous function)

= Lt _x→0 [log (1+f(x) - 1)/f(x) - 1] Lt _x→0(f(x) - 1)g(x) (as both limits are finite)

= Lt _{f(x)-1→0} [log (1+f(x) - 1)/f(x) - 1] Lt _x→0(f(x) - 1)g(x) = Lt _x→0(f(x) - 1)g(x)

Hence Lt _x→0 y = e^{Lt _x→0(f(x) - 1)g(x)}

thanx prophet for another method

bhaiyya why isnt this pinked

This one can be written as Lt_x→0 (1+2x)^1/x (1+x)^1/x

Lt_x→0 (1+x)^1/x = e and Lt_x→0 (1+2x)^1/x = [Lt_2x→0 (1+2x)^1/2x]² = e²

So the limit is e³

bhaiya i see it that complete proof is not necessary

but ur proof doesnt solve this problem either

or does it

i understand ur steps

but how does it prove the initial statement

some one please reply

e³

guyz i know this is a silly question

but plz give me the proof for that

how do u get the e^(Ltx→0f(x)-1)g(x) thing

oops... i'm late... [6]

If Lt x→0 f(x)^g(x) is of the form 1^∞

then, the given limit is equivalent to e^{Lt x→0 [f(x)-1].g(x)}

Solve using above concept...

the answer will be e³

[1]

e^(f(x)-1)g(x)
for lt_x→0 f(x)^g(x)
where f(x)→1 and g(x) → ∞ when x→0

can u pls show ur method