1okie
lets leave this part to prophet then :) we enjoy reading his posts
(though we don't understand much) [1]
11thanx a lot for the proof
341And about why Lt x→0 f(x)g(x) ≡ eLt 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 = eLt x→0(f(x) - 1)g(x)
11
Subash
·2009-01-15 09:22:52
thanx prophet for another method
bhaiyya why isnt this pinked
341
Hari Shankar
·2009-01-15 07:41:19
This one can be written as Ltx→0 (1+2x)1/x (1+x)1/x
Ltx→0 (1+x)1/x = e and Ltx→0 (1+2x)1/x = [Lt2x→0 (1+2x)1/2x]2 = e2
So the limit is e3
11
Subash
·2009-01-15 06:10:13
bhaiya i see it that complete proof is not necessary
but ur proof doesnt solve this problem either
or does it
11
Subash
·2009-01-15 05:41:10
i understand ur steps
but how does it prove the initial statement
62there is no formula... to learn...
but if u are nto able to understand ti.. then learn it :)
there is no problem in learning this one..
but u should see th eproof from my last post
11
Subash
·2009-01-15 05:34:39
btw pink is better than green solutions are beautiful
11
Subash
·2009-01-15 05:13:12
shud i just remember the formula for tis one
62
Lokesh Verma
·2009-01-15 05:03:14
haha..
I sometimes start to wonder if he is still preparing for JEE... :)
62
Lokesh Verma
·2009-01-15 04:54:16
it is provable i believe and comes from that..
if u want i can try to give the proof but trust me there is no need to go that deep..
among our "active" users i think only prophet understands that part well enuf... (not that it is bad to know.. but at this stage there are better things to know!)
62
Lokesh Verma
·2009-01-15 04:53:00
well philip .. that is provable.. but it needs a bit of pure mathematics...
You know the most basic definiton the epsilon delta definiton of existance of limit?
1
Philip Calvert
·2009-01-15 04:39:42
[7][7][7]
[7][7][7]
clueless how does this prove e^(Ltx→0f(x)-1)g(x)
11
Subash
·2009-01-15 04:01:04
guyz i know this is a silly question
but plz give me the proof for that
3i am also getting e3
mtd: take log and use l'hopital rule
11
Subash
·2009-01-15 03:46:42
how do u get the e^(Ltx→0f(x)-1)g(x) thing
13If Lt x→0 f(x)g(x) is of the form 1∞
then, the given limit is equivalent to eLt x→0 [f(x)-1].g(x)
Solve using above concept...
the answer will be e3
[1]
1
Philip Calvert
·2009-01-15 03:38:34
e(f(x)-1)g(x)
for ltx→0 f(x)g(x)
where f(x)→1 and g(x) → ∞ when x→0
