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Let `p=(lim)_(xvec0+)(1+tan^2sqrt(x))^(1//2x)`then `logp`is equal to:(1) 2(2) 1(3) `1/2`(4)`1/4` |
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Answer» `p = lim_(x->0^+) ( 1+ tan^2 sqrtx)^(1/(2x))` `log p = lim_(x->0^+) log(1+ tan^2 sqrtx)^(1/(2x))` `= lim_(x->0^+) 1/(2x) log(1+ tan^2 sqrtx)` `= lim_(x->0^+) (Log(1+tan^2 sqrt x))/(2x)` using L hospital rule `lim_(x->0) (ax^2)/(bx^3) = (2ax)/(3bx^2)` `= lim_(x->0^+) (1/(1+ tan^2 sqrtx) xx 2 tan sqrt xx sec^2 sqrt x xx 1/(2 sqrt x))/2` `log p = lim_(x->0+) ( tan sqrt x xx sec^2 sqrt x)/ ( 2 sqrt x xx ( 1 + tan^2 sqrtx))` `= (1 xx 1)/(2 xx (1+0))` `log p = 1/2 ` option 2 is correctAnswer |
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