Question

(i) The value of lim, _π/2 sinx^tanx​

Answer 1

limx→π/2(sinx)tanx

 

Since sinx and tanx are continuous functions, using the continuity of ex, this expression has the equivalent form:

limx→π/2elog((sinx)tanx)

log(sinx)tanx=tanxlog(sinx)=log(sinx)1tanx

Taking the limit of this fraction as x goes to π/2 has the indeterminate form of −∞/∞, so we can apply Hôpital's rule.

limx→π/2log(sinx)1tanx=1sinx(−cosx)−1sin2x=11(0)−11=0

⟹limx→π/2(sinx)tanx=e0=1

Answer 2

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