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Differential Calculus

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16 Aug 2007 10:26:54 IST

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limit problem

None

LIM ( ln ( {x} + | [x] | ) ) ^ {x}

x-->0

where [x] == greatest integer function & {X} == fractional part of x.

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Comments (7)

shashidhar sundareisan

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16 Aug 2007 11:31:54 IST

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when x tends to 0⁺
[x] is 0.
{x} behaves as x.
hence limit becomes - infinity raised to 0 form.
by slight manipulation n l hospital rule we get rhl =1.
when x tends to 0^-
[x] is -1
{x} behaves as 1 - mod(x).
hence lhl = ln 2.
hence limit doesnt exist.
pls tell me if d ans is wrong.

venkat v.t

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18 Aug 2007 22:25:19 IST

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Dear Shashidhar LHL correct.But while doing RHL you got base
-infinity .Base should not be negative for exponential function.So we should find only LHL.Therefore,the limit is ln2.

deepak_agarwal

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8 Sep 2007 12:14:46 IST

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well a really gud problem...i actually appreciate both shashidhar and venkat..shashidhar for actually following the rite approach and venkat for pointing out small error...actually when u have -infinity to the power of 0 ...u take log both sides but unfortunately here it results in log of -infinity..so it does nt exist

venkat v.t

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9 Sep 2007 17:59:05 IST

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Dear Deepak!

Will u please say the limit of root(x) when x tends to 0?

kanika gupta

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12 Sep 2007 21:19:50 IST

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venkat the expression given in x is not base of log & shashidhars LHL will be ln2^0 which will make it 1

Ambareesh Rishi

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12 Sep 2007 21:47:45 IST

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@venkat
limit of root x when x tending to 0 doesnt exists

root can never be negative and in this case lhl is imaginary

siva viper

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13 Sep 2007 01:13:08 IST

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yeah venkat
for root(x) the rhl exists but lhl doesn''t e xist.........therefore the limit doen't exist.........its simple buddy........noe could anyone please explain the lhl partr in the given prob. by aiimiit...............plzzzzzzzzzz

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