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Question
If `f(x)=[tan(pi/4+x)]^(1/x), `
= k ,for x=0
is continuous at x=0 , find k.
Solution
`f(x)=[tan(pi/4+x)]^(1/x), " for "x!=0`
`f(0)=k`
Since f(x) is continuos at x=0
`lim_(x->0)f(x)=f(0)`
`lim_(x->0)[tan(pi/4+x)]^(1/x)=k`
`lim_(x->0)[(1+tanx)/(1-tanx)]^(1/x)=k`
`lim_(x->0)[1+(1+tanx)/(1-tanx)-1]^(1/x)=k`
`lim_(x->0)[1+(1+tanx-1+tanx)/(1-tanx)]^(1/x)=k`
`lim_(x->0)[1+(2tanx)/(1-tanx)]^(1/x)=k`
`lim_(x->0)[1+(2tanx)/(1-tanx)]^(1/((2tanx)/(1-tanx))xx((2tanx)/(x.(1-tanx))))=k`
`e^(lim_(x->0)(2tanx)/(x.(1-tanx)))=k {becauselim_(x->0)[1+x]^(1/x)=e}`
`e^(2lim_(x->0)(tanx)/(x)xxlim_(x->0)1/(1-tanx))=k {becauselim_(x->0)[tanx/x]=1}`
`e^(2xx1xx1/(1-0))=k`
`k=e^2`
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