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If the functionf(x) = [tan(pi/4 + x)]^(1/x) for x ne 0 is = K for x = 0 |
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Answer» e Since, f(x) is continuousat `x = 0`,then `f(0) = underset(xrarr0)("LIM")f(x)` `= underset(xrarr0)("lim") [tan((x)/(4) + x)]^(1//x)` `RARR K = underset(xrarr0)("lim") [(1+TANX)/(1-tanx)]^(1/x)` [`1^(oo)` form] `=e^(underset(xrarr0)("lim")[(1+tanx)/(1-tanx)-1]1/x)` `=e^(underset(xrarr0)("lim")((2tanx)/(1-tanx)-1)1/x)` `=e^(underset(xrarr0)("2lim")(tanx)/(x)underset(xrarr0)("lim")(1)/(1-tanx))` `[ :' underset(xrarr0)("lim")(tanx)/(x) = 1]` `:. K = e^(2.1(1/1-0)) = e^(2)` |
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