IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

newkid

_{[x ]}

_{[0 ]} (1 - sinx)^1/(-sinx)

is it 1form or 1^-form ???

can the formula

_{[x ]}

_{[0 ]} f(x)^g(x).........(such that_{[x ]}

_{[0 ]} f(x) = 1 and _{[x ]}

_{[0 ]} g(x) =

)

= e^ [_{[x ]}

_{[0 ]} g(x).{f(x) - 1}] be used here ??

bhuwanaroracorroded

it is -

here

and that formula can b used here

newkid

1^ (-infinity) = 0 na???

kyun bhai???

to phir is ka answer to zero ho jayega... then why to use formula

neeraj_agarwal_1990

nopes...
1^ (-infinity) = 0 is wrong...
in fact its (tending to 1)^ (tending to -infinity) which is an indeterminate form...u can check bu taking log....

panks_01

hey newkid

1 ^ infinity is not equal to zero

it is an unknown form

n dis type of ques can be solved by using d foll formula:

_{[ x]}

_{[a ]} f(x)^g(x) ........where x

a, f(x)

1, g(x)

= e ^ [ _{[ x]}

_{[a ]} g(x) { f(x) - 1}]

ya u cn use dis formula here

plz rate if corr.........

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