IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

nunknown91

_{[x ]}

_[0]|x|^sinx is ....

spideyunlimited

limit is equal to 1.
reason --> |x| when x tends to 0 is a very small number tending to 0 but not zero and sinx for very small values of x is also equal to x so it is a very small value tending to 0.
a very small number tending to a power very close to 0..is equal to 1...
(0.0000001)^(0.0000001) tends towards (0.00000001)^0 = 1

ayshwarya

its 1 dude v must cnsdr 0 as critical pts by taking LHL n RHL v may get Lt (-h) ^sinh=e^[(-h-1) sinh ] =1 n in same way d oder v get as
h-0

Lt hsinh =e^[(h-1)sinh]=1
h-0 rate me if u like it

venkat_tatolu

Base is positive ( since modx positive but not zero when x tends to zero) and the index is 0 (limit)

So some thing power zero is 1.

So limit is 1

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