simple

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)}

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

btw pink is better than green solutions are beautiful

some one please reply

e³

[7][7][7]
[7][7][7]
clueless how does this prove e^(Ltx→0f(x)-1)g(x)

guyz i know this is a silly question

but plz give me the proof for that

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

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