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underset( x rarr 0 ) ("lim") ( cos mx )...

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

A

`e^(-m^(2) n //4)`

B

`e^(-m^(2) n //2)`

C

` e^(-m n^(2) //2)`

D

`e^(-m n^(2) //4)`

Text Solution

Verified by Experts

The correct Answer is:
B
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MOTION-LIMIT-EXERCISE -1
  1. The limit underset( x rarr a ) ("Lim") (2 -( a)/(x))^(tan ((pix)/( 2a)...

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

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

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

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

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

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

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

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

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  10. underset( x rarr 1 )("lim")( root ( 3)(7+x^(3))- sqrt( 3 +x^(2)))/( x-...

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  11. ABC is an isosceles triangle described in a circle of radius r.If AB=...

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  12. The figure shows a right triangle with its hypotenuse OB along the y-a...

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  13. lim(n->oo){(n/(n+1))^alpha+sin(1/n)}^n where alpha epsilon Q is equal ...

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  14. underset( x rarr 0 ) ( "lim")( x tan 2x - 2x ta n x )/(( 1- cos 2x)^(...

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  15. If underset( x rarr0 )("lim")((sin nx) [ ( a-n ) nx - 2 tan x])/( x^(2...

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  16. lim[n->oo][5^[n+1]+3^n-2^[2n]]/[5^n+2^n+3^[2n+3]]

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  17. the value of lim(x->0)(cos(sinx)-cosx)/x^4 is equal to:

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  18. lim[x->0][e^[-x^2/2]-cosx]/[x^3sinx]

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  19. If underset( x rarr 0 )( "lim") ( e^(-nx) + e^(nx) -cos""( nx)/( 2) - ...

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  20. underset( x rarr 2) ( "lim") ( sin ( e^(x-2) - 1))/( ln ( x-1))=

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