If `lim_(x to 0) (cos4x+acos2x+b)/(x^(4))` is finite, find a and b using expansion formula. Also, find the limit.

If `lim_(x to 0) (cos4x+acos2x+b)/(x^(4))` is finite, find a and b usi

1.	If `lim_(x to 0) (cos4x+acos2x+b)/(x^(4))` is finite, find a and b using expansion formula. Also, find the limit.
Answer» `underset(xto0)lim(cos4x+acos2x+b)/(x^(4))="Finite"` Using expansion formula for `cos4x" and "cos2x,` we get `underset(xto0)lim((1-(4x)^(2)/(2!)+(4x)^(4)/(4!))+a(1-(2x)^(2)/(2!)+(2x)^(4)/(4!))+b)/(x^(4))` or `" "underset(xto0)lim((1+a+b)+(-8-2a)x^(2)+((32)/(3)+2/3a)x^(4)+...)/(x^(4))` or `" "1+a+b=0` `-8-2=0` Solving (1) and (2) for a and b, we get `a=-4" and " b=3` Also, Limit`=32/3+2/3a=(32-8)/(3)=8`

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`lim_(xto oo) ((nsqrt(a)+nsqrt(b))/(2))^n,a,b,gt 0` equalsA. 1B. `sqrt(pq)`C. `pq`D. `(pq)/(2)`
If `lim_(xto oo){(x^2+1)/(x+1)-(ax+b)}to oo`, thenA. `a in (1,oo)`B. `a ne 1 , b in R`C. `a in (-oo,1)`D. none of these
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\(f\left(x\right)=\left[\left(\frac{min\left(t^2+4t+6\right)\sin \left(x\right)}{x}\right)\right]\)where [.] represents greatest integer function, then find \(\lim _{x\to 0}\left(f\left(x\right)\right)\)
`lim_(xrarr0) ((1-cos2x)^2)/(2xtanx-xtan2x)` , isA. 2B. `-(1)/(2)`C. `(1)/(2)`D. `-2`
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`lim_(xrarroo) ((x+2)/(x+1))^(x+3)` is equal toA. 1B. eC. `e^2`D. `e^3`
`lim_(x->0)((1-cos2x)sin5x)/(x^2sin3x)`A. `(10)/(3)`B. `(3)/(10)`C. `(6)/(5)`D. `(5)/(6)`
`lim_(x->0)((1-cos2x)sin5x)/(x^2sin3x)`A. `10//3`B. `3//10`C. `6//5`D. `5//6`