Evaluate `lim_(xto(pi)/(6)) (2-sqrt(3)cosx-sinx)/((6x-pi)^(2)).`

1.	Evaluate `lim_(xto(pi)/(6)) (2-sqrt(3)cosx-sinx)/((6x-pi)^(2)).`
Answer» `underset(xto(pi)/(6))lim(2-sqrt(3)cosx-sinx)/((6x-pi)^(2))" "`(0/0 from) `=underset(xto(pi)/(6))lim2xx(1-((sqrt(3))/(2)cosx+1/2sinx))/((6x-pi)^(2))` `=underset(xto(pi)/(6))lim2xx(1-cos(x-(pi)/(6)))/(36(x-(pi)/(6))^(2))` `=underset(xto(pi)/(6))lim(2sin^(2)(x/2-pi/12))/(18xx4(x/2-pi/12)^(2))` `=(1)/(36)(underset(xto(pi)/(6))lim(sin(x/2-pi/12))/((x/2-pi/12)))^(2)=(1)/(36)`

Discussion

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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
Evaluate:\(\lim\limits_{x \to 1}\frac{\sqrt {1+x}-\sqrt {1-x}}{1+x}\)
\(f\left(x\right)=\left[\left(\frac{min\left(t^2+4t+6\right)\sin \left(x\right)}{x}\right)\right]\)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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