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The value of limit x tend to 0 ((sinx)^1/x + (1/x)^sinx) equals

The value of limit x tend to 0 ((sinx)^1/x + (1/x)^sinx) equals

Grade:12th pass

1 Answers

Aditya Gupta
2081 Points
5 years ago
obviously limit x tend to 0 (sinx)^1/x=0 because sinx is less than one and when it is raised to power of 1/x (which tends to infinity), it approaches zero.
now, limit x tend to 0 (1/x)^sinx=limit x tend to 0 1/x^sinx= 1/limit x tend to 0 x^sinx
to find limit x tend to 0 x^sinx, take log and we get
sinx*logx=(sinx/x)*x*logx
since sinx/x=1 as x is near zero, so we are left with
xlogx. now put x=1/y so that y tends to infinity.
so limit becomes
-logy/y
apply lhospital rule
-1/y=0 as y tends to inf.
take antilog
L=1

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