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Differential Calculus

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14 Sep 2007 12:42:38 IST

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550

limit

None

lt x tends to 0

(1+sinx)^1/x^2

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Mahesh Jain

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14 Sep 2007 12:44:50 IST

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the answer is e by multiplying the power by sinx and dividing by sinx then take limits of power and limits (1+sinx)^1/sinx=e

chandrashekhar shekhar

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14 Sep 2007 12:49:42 IST

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what is the answer???

Mahesh Jain

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14 Sep 2007 12:52:34 IST

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chandrashekhar shekhar

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14 Sep 2007 12:55:46 IST

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sorry ur answer is wrong

Mahesh Jain

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14 Sep 2007 12:56:41 IST

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so wts the answer

sri hari

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14 Sep 2007 12:57:29 IST

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i thick the ques,,,...is

lt x tends to 0

(1+sinx)^1/x

is it

manu kumar

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Joined: 20 Mar 2007

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14 Sep 2007 12:59:23 IST

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it is infinity

bcoz given limit = lt _x->0(1+sin x)^{ (1/sin x)*(sin(x)/x)*1/x } = lt _x->0e^1/x= infinity

chandrashekhar shekhar

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14 Sep 2007 13:01:31 IST

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question is absoluetly right
and answer is limit does not exists

chandrashekhar shekhar

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14 Sep 2007 13:02:34 IST

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now i have given u answer also

manu kumar

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14 Sep 2007 13:04:15 IST

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then is my explanation right ?????????????

chandrashekhar shekhar

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14 Sep 2007 13:04:22 IST

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lt x tends to 0
(1+sinx)^(1/(x^2))

it is from fiitjee

chandrashekhar shekhar

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14 Sep 2007 13:05:50 IST

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no explanation is not right
as infinity does not mean limit does not exists

Mahesh Jain

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14 Sep 2007 13:08:33 IST

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hey is my explanation right or not tell me

chandrashekhar shekhar

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14 Sep 2007 13:12:16 IST

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sorry bro
because
e^1/x
when approached left side it is zero
other side infinity
limit not exist at x=0

try other limit question that i have given

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