If `lim_(xto1) (2-x+a[x-1]+b[1+x])` exists, then a and b can take the values (where `[.]` denotes the greatest integer function)A. is always equal to -1B. is always equal to +1C. does not exist None of theseD.

If `lim_(xto1) (2-x+a[x-1]+b[1+x])` exists, then a and b can take the

1.	If `lim_(xto1) (2-x+a[x-1]+b[1+x])` exists, then a and b can take the values (where `[.]` denotes the greatest integer function)A. is always equal to -1B. is always equal to +1C. does not exist None of theseD.
Answer» Correct Answer - B::C Since the greatest integer function is discountinuous at integral values of x, for a given limit to exist both left-and right-hand limits must be equal. `L.H.L.=underset(xto1^(-))lim(2-x+a[x-1]+b[1+x])` `=2-1+a(-1)+b(1)=1-a+b` `R.H.L.=underset(xto1^(+))lim(2-x+a[x-1]+b[1+x])` `=2-1+a(0)+b(2)=1+2b` On comparing, we have `-a=b`.

`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`
The value of `lim_(xrarroo) ((x+3)/(x-1))^(x+1)` isA. eB. `e^2`C. `e^4`D. `1//e`
`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`