Evaluate `lim_(ntooo) {cos((x)/(2))cos((x)/(4))cos((x)/(8))...cos((x)/(2^(n)))}`.

Evaluate `lim_(ntooo) {cos((x)/(2))cos((x)/(4))cos((x)/(8))...cos((x)/

1.	Evaluate `lim_(ntooo) {cos((x)/(2))cos((x)/(4))cos((x)/(8))...cos((x)/(2^(n)))}`.
Answer» We know that `costhetacos2thetacos4theta...cos2^(n-1)theta=(sin2^(n)theta)/(2^(n)sintheta)` `:." "cos((x)/(2))cos((x)/(4))cos((x)/(8))...cos((x)/(2^(n)))=(sinx)/(2^(n)"sin"((x)/(2^(n))))` `:." ""Given limits is "underset(ntooo)lim(sinx)/(2^(n)sin((x)/(2^(n))))=underset(ntooo)lim((sinx)/(x)((x)/(2^(n))))/(sin((x)/(2^(n))))` `=(sinx)/(x)`

`lim_(xto oo) (1)/(n) {1+e^(1//n)+e^(2//n)+e^(3//n)+.....+e^((n-1)/(n))}` is equal toA. eB. `-e`C. `e-1`D. `1-e`
`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 \(\lim _{x\to 0}\left(f\left(x\right)\right)\)
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`lim_(x->0)((1-cos2x)sin5x)/(x^2sin3x)`A. `(10)/(3)`B. `(3)/(10)`C. `(6)/(5)`D. `(5)/(6)`
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