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lim(x->pi/2) ((1-tan(x/2))(1-sinx))/((1...

`lim_(x->pi/2) ((1-tan(x/2))(1-sinx))/((1+tan(x/2))((pi-2x)^3))`

A

`1//16`

B

`-1//16`

C

`1//32`

D

`-1//32`

Text Solution

Verified by Experts

The correct Answer is:
C
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MOTION-LIMIT-EXERCISE -1
  1. lim(x->oo)(x^2sin(1/x)+x+1)/(x^2+x+1) is equal to

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  2. Evaluate: ("lim")(nvecoo)ncos(pi/(4n))sin(pi/(4n)) where (x , y)vec(0...

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  3. lim(x->pi/2) ((1-tan(x/2))(1-sinx))/((1+tan(x/2))((pi-2x)^3))

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  4. The limit underset( x rarr0 ) ( "lim") ( 1- cos x sqrt( cos 2x))/( x^(...

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  5. lim(x->0^+)cos^(- 1)(1-x)/(sqrt(x))

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  6. underset( x rarr 0 ) ("lim") [ ( 1-e^(x)) ( sin x )/( |x|)] is equal ...

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  7. Evaluate: underset(hrarr0)("lim")(2[sqrt(3)sin(pi/6+h)-cos(pi/6+h)])/(...

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  8. lim(x->0) sin(6x^2)/(lncos(2x^2-x)

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  9. The value of underset( x rarr 0 ) ( "Lim") ( sin ( ln ( 1+ x)))/( ln (...

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  10. underset( x rarr pi //2) ("Lim") ( 2 ^(cos x) - 1)/( x ( x-pi //2))= ...

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  11. lim(x->0)((4^x-1)^3)/(sin(x/p)ln(1+(x^2)/3) is equal to

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  12. The limit underset( x rarr a ) ("Lim") (2 -( a)/(x))^(tan ((pix)/( 2a)...

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  13. underset( x rarr oo) ( "Lim") ((x+ 2)/( x-2))^(x+1)=

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  14. The value of underset( x rarr pi //4) ("Lim") ( 1+[x])^(1//ln( tan x)...

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  15. lim(x->oo) ((x^2-2x+1)/(x^2-4x+2))^x is equal to

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  16. underset( x rarr 0 ) ("lim") ( cos mx )^(n//x^(2))

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  17. If lim(x -> oo) (1 + a/x + b/x^2)^(2x)= e^2 then the values of a and ...

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  18. underset( x rarr0 ) ( "lim" )( cot ((pi)/( 4) + x ))^("cosecx")=

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  19. lim(x->oo)(sin(1/x)+cos(1/x))^x

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  20. lim(x->oo) (root(3)(1+x)-root(3)(1-x))/x

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