Evaluate `lim_(x to 0) x[tan^(-1)((x+1)/(x+2))-tan^(-1)((x)/(x+2))].`

1.	Evaluate `lim_(x to 0) x[tan^(-1)((x+1)/(x+2))-tan^(-1)((x)/(x+2))].`
Answer» `underset(xtooo)limx["tan"^(-1)(x+1)/(x+2)-"tan"^(-1)(x)/(x+2)]` `=underset(xtooo)limxtan^(-1)(((x+1)/(x+2)-(x)/(x+2))/(1+(x+1)/(x+2).(x)/(x+2)))` `=underset(xtooo)limxtan^(-1)((x+2)/(2x^(2)+5x+4))` `=underset(xtooo)lim(tan^(-)((x+2)/(2x^(2)+5x+4))/((x+2)/(2x^(2)+5x+4)))xx(x(x+2))/(2x^(2)+5x+4)` `=1xx1/2=1/2`

Discussion

`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
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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`lim_(xrarroo) ((x+2)/(x+1))^(x+3)` is equal toA. 1B. eC. `e^2`D. `e^3`
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`lim_(x->0)((1-cos2x)sin5x)/(x^2sin3x)`A. `10//3`B. `3//10`C. `6//5`D. `5//6`