InterviewSolution
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InterviewSolution
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`lim_(x-pi/2) (cot x - cosx)/(pi-2x)` equals: (1) `1/8` (2) `1/4` (3) `1/24` (4) `1/16` |
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Answer» Let `x = pi/2 - h`, Then, our limit becomes, `Lim_(h->0) (cot(pi/2-h) - cos(pi/2-h))/(pi -(2(pi/2-h))^3 )` `=Lim_(h->0) (tanh - sinh)/(pi -2(pi/2-h))^3 ` `=Lim_(h->0) (sinh(1-cosh)/(cosh))/(8h^3) ` `=Lim_(h->0) (tanh(2sin^2(h/2)))/(8h^3) ` `=Lim_(h->0) (tanh(sin^2(h/2)))/(4h^3) ` `=Lim_(h->0) (tanh/h)(sin^2(h/2))/(4*4(h/2)^2) ` `=1*1/16*1 = 1/16` `:. ` Option `(4)` is the correct option. |
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