If ` a .lim_(xrarr 1) x^(1//1-x)+b=e^(-1)(a ge 1,bge 0)` , thenA. `a=1,b=e^-1`B. `a=2,b=e^-1`C. `a=-1,b=e^-1`D. `a=1,b=0`

If ` a .lim_(xrarr 1) x^(1//1-x)+b=e^(-1)(a ge 1,bge 0)` , thenA. `a=1

1.	If ` a .lim_(xrarr 1) x^(1//1-x)+b=e^(-1)(a ge 1,bge 0)` , thenA. `a=1,b=e^-1`B. `a=2,b=e^-1`C. `a=-1,b=e^-1`D. `a=1,b=0`
Answer» Correct Answer - D We have `a. lim_(xto1) x^(1//1-x+b=e^(-1))` ` rArr a. lim_(xto1)[1+(x-1)]^(1//1-x)+b=e^(-1)` ` rArr ae ^(lim_(xto1)(x-1)/(1-x))+b=e^(-1)` ` rArr a.e ^-1+b=e^-1rArr a=1and b=0`.

`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`