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waise maine bhi to 9/5 hi kaha tha .... [3] (-1/5)+2
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edited there [2]
i want to check my ans for the question
limx→∞((x-1)(x-2)(x-3)(x+5)(x+10))1/5-x is
more hint
what is
limit x going to infinity ( x5-1 )1/5-x
limit x going to infinity ( x5-a x4 )1/5-x
lets take this
(x5+ax4+....+c)1/5-x
=x {(1+a/x+....+c/x5)1/5-1}
we know that
(1+t)1/5 = 1+1/5t+.... higher powers ...
hence the above will be similar to
x{(1+a/5x+... higher orders of 1/x!)-1}
hence this will be a/5....
when x goes to infinity...
here in the above question, a=-9
hence.. -9/5
[3][3][3][3]
waise maine bhi to 9/5 hi kaha tha .... [3] (-1/5)+2
[6]
edited there [2]
Curses be on me...
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are wahi hai [6]
galti ho jata hai...
mere par mat haso.. [2]
sabse pahle maine hi ye kaha tha ....
http://targetiit.com/iit_jee_forum/posts/limits_387.html
Guys, I don't understand a thing of what is going on here from the beginning [2]
Pls post your complete thought process along with your replies, guys. Please.
If i were to make a tukka for this one!
my tukka says
(-1-2-3+5+10)/5 = 9/5
[2]
don't laugh on my mistake in tukka atleast
Nishant Bhaiya:(It is not very very simple.. but doable!)
but i did it in 10 secs without any pen... with my tukka....
yipieeeeeee [4]
same question here also :
and my same tukka there also [3]
http://targetiit.com/iit_jee_forum/posts/limits_387.html
typing mistake..
thats (-1+10)/5 [3]
i.e. (-1/5)+2 [3]
it for sure (-1/5)+2.. [3]
with my evergreen tukka...
:P
{ (x-1)(x-1)(x-1)(x-1)(x-1) }1/5 - x
it is -1 ?
so taht should be 10
take another example
{ (x-1)(x-1)(x-1)(x-1)(x-1) }1/5 - x
what will it be ??
If this is not zero.. then can u justify it for another arbitrary polynomial?
do u realise how to solve the previous question that i posted!?
then u may be able to get the thought process going!
(It is not very very simple.. but doable!)
(x-1)(x-2)(x-3)(x+5)(x+10) this term is equal to x5 when x approaches to infinity so that the exp reduces to limx→∞(x-x) which is zero. please correct me.
areh sir that one is easy
bring that into denominator by multiplying with came term with + in between