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L =lim(x->0)((1)/(ln(1+x))-1/(ln(x+sqrt(...

`L =lim_(x->0)((1)/(ln(1+x))-1/(ln(x+sqrt(1+x^2))))` then find the value of `(L+153)/L`.

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MOTION-LIMIT-EXERCISE-3
  1. lim(x->oo)((x+c)/(x-c))^x= 4 then find c.

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  2. Evaluate: lim(x->oo) { ((a1)^(1/x)+(a2)^(1/x)+.... +(an)^(1/x))/n}^(n...

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  3. L =lim(x->0)((1)/(ln(1+x))-1/(ln(x+sqrt(1+x^2)))) then find the value ...

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  4. underset( x rarr 0 ) ( "Lim") ((x- 1+ cos x )/( x))^((1)/( x))

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  5. underset( x rarr oo)("Lim")[cos (2pi ((x)/(1+ x ))^(a))]^(x^(2))a in Q

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  6. lim(n->oo) ((sqrt(n^2+n)-1)/n)^(2sqrt(n^2+n)-1)

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  7. underset( x rarr 0 ) ("Lim") [ ((1+x)^(1//x))/( e ) ]^(1//x)

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  8. underset( x rarr oo) ("Lim") ((cos h ( pi //x))/(cos( pi//x)))^(x^(2))

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  9. Evaluate lim(x->1)(root13 x- root 7 x)/(root5 x- root3 x)

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  10. lim[x->1] ([sum[k=1]^100x^k]-100])/(x-1)

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  11. lim(x ->oo)x^(2)(sqrt((x+2)/(x))-root(3)((x+3)/(x))) equals

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  12. A circular are of radius 1 subtends an angle of x radians 0 < x < pi/2...

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  13. A circular are of radius 1 subtends an angle of x radians 0 < x < pi/2...

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  14. A circular are of radius 1 subtends an angle of x radians 0 < x < pi/2...

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  15. At the end points A,B of the fixed segment of length L, lines are draw...

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  16. At the endpoint and midpoint of a circular are AB, tangent lines are ...

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  17. lim(x->0) [ ln(1+sin^2x) cot ln^2(1+x)]

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  18. lim(x->oo)(a(2x^3-x^2)+b(x^3+5x^2-1)-c(3x^3+x^2))/(2(5x^4-x)-b x^4+c(4...

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  19. find a&b if lim(x->oo) ((x^2+1)/(x+1)-ax-b)=0

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  20. lim(x rarr oo) (sqrt(x^2-x+1)-a x-b)=0, then a+b=

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