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Lim(x->0) {x(1+acosx)-bsinx}/x^3=1 then...

`Lim_(x->0) {x(1+acosx)-bsinx}/x^3=1` then

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Evaluate find a and b if \ ("Lim")_(x->0)(x(1+acosx)-bsinx)/(x^3)=1

if l=lim_(x->0) (x(1+acosx) - bsinx)/x^3 = lim_(x->0) (1+acosx)/x^2-lim_(x->0) (b sinx)/x^3 where l in R , then

if l=lim_(x->0) (x(1+acosx) - bsinx)/x^3 = lim_(x->0) (1+acosx)/x^2-lim_(x->0) (b sinx)/x^3 where l in R , then

F(x) is the function such that (lim)_(x->0)(f(x))/x=1\ a n d\ (lim)_(x->0)(x(1+acosx))/((f(x))^3)=1 , then find the value of a

lim_(x rarr0)(x(1+a cos x)-b sin x)/(x^(3))=1 then

Let lim_(x rarr0)(f(x))/(x)=1 and lim_(x rarr0)(x(1+p cos x)-q sin x)/((f(x))^(3))=1 then

lim_(x->0) (3x+1)/(x+3)

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If lim_(x rarr0)(x(1+mecx)-n sin x)/(x^(3))=1 then

lim_(x to 0) (x^3 sin(1/x))/sinx

A DAS GUPTA-Limit, Indetermine Form-Exercise
  1. ("lim")(xvec0)((1+5x^2)/(1+3x^2))^1//x^2=

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  2. Fill in the blanks: lim(x rarr oo) ((x^2+1)/(x^2-1))^(x^2)=

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  3. Lim(x->0) {x(1+acosx)-bsinx}/x^3=1 then

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  4. If f'(2)=2, f''(2) =1, then lim(xto2)(2x^2-4f'(x))/(x-2), is

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  5. If f(9)=9 , f'(9)= 4 then lim (x rarr9) (sqrt f(x) -3)/(sqrt(x)-3) =.

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  6. If G(x)=-sqrt(25-x^2),t h e n("lim")(xvec1)(G(x)-G(1))/(x-1)i s 1/(24...

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  7. If f(a) =2, f'(a)=1, g(a) =-1 , g' (a)=2, then the value of lim(xto a)...

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  8. If lim(x rarr a)(px+q)/(qx+p)=l and lim(x rarr0)(px+q)/(qx+p)=m where ...

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  9. If f(x)=sqrt((x-sinx)/(x+cos^2x)), then lim (x->oo)f(x) is

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  10. lim(x->0)(x e^x-log(1+x))/(x^2) equals

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  11. Lim(x->0) {x(1+acosx)-bsinx}/x^3=1 then

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  12. Let f(x) be a twice-differentiable function and f"(0)=2. The evaluate:...

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  13. lim (x rarr pi/2) (2x-pi)/cos x is equal to

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  14. evaluate lim(thetararrpi/4)(sqrt2-costheta-sintheta)/(4theta-pi)^2

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  15. If f(x)=cot^(-1) ((3x-x^3)/(1-3x^2)) and g(x)=cos^(-1)((1-x^2)/(1+x^2)...

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  16. 23. The value of lim(h->0)(1/(h(8+h)^(1/3)) - 1/(2h)) equals

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  17. Evaluate: ("lim")(hvec0)(2[sqrt(3)sin(pi/6+h)-cos(pi/6+h)])/(sqrt(3)h(...

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  18. lim(x->0)(sqrt((1-cos2x)/2))/x

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  19. lim(x rarr 0) (log(1+x))/(3^x-1)=1/(log(e)(3))

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  20. lim(x->a)(cosx-cosa)/(cotx-cota)=

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