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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: limit
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nunknown91 (66)

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[x ][0 ] (1/x)sinx

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akku (1142)

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LIMIT DOES NOT EXIST
only right hand limit exists
rhl
y=lim x--->0+ (1/x)^sinx
logy= -lim x-->0sinx logx
      =-lim x-->0 logx/cosec x  (inf/inf form)
       by LH
      = lim x-->0tanx/x *sinx=0
y=e^0
so RHL=y=1
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nunknown91 (66)

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why did u apply lhospital in lh  limit and y not in rhl


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akku (1142)

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the left hand limit doesnot exist as log x is defined only fr x belonging to +ve real so i applied it to rhl
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nunknown91 (66)

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thanq....

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