If `f(x)=(3x^2+a x+a+1)/(x^2+x-2),`then which of the following can be correct`("lim")_(xvec1)f(x)e xi s t s a=-2``("lim")_(xvec-2)f(x)e xi s t s a=13``("lim")_(xvec1)f(x)=4/3``("lim")_(xvec-2)f(x)=-1/3`A. `"ln "a_(1)`B. `e^(a_(n))`C. a_(1)`D. `a_(n)`

If `f(x)=(3x^2+a x+a+1)/(x^2+x-2),`then which of the following can be

1.	If `f(x)=(3x^2+a x+a+1)/(x^2+x-2),`then which of the following can be correct`("lim")_(xvec1)f(x)e xi s t s a=-2``("lim")_(xvec-2)f(x)e xi s t s a=13``("lim")_(xvec1)f(x)=4/3``("lim")_(xvec-2)f(x)=-1/3`A. `"ln "a_(1)`B. `e^(a_(n))`C. a_(1)`D. `a_(n)`
Answer» Correct Answer - A::B::C::D `f(x)=(3x^(2)+ax+a+1)/((x+2)(x-1))` As `xto1,D^(r)to0," Hence as "xto 1,N^(r)to0.` Therefore, `3+2a+1=0" or "a=-2` As `xto-2,D^(r)to0." Hence as "xto-2,N^(r)to0.` Therefore, `12-2a+a+1=0" or "a=13` Now, `underset(xto1)limf(x)=underset(xto1)lim(3x^(2)-2x-1)/((x+2)(x-1))=underset(xto1)lim((3x+1)(x-1))/((x+2)(x-1))=(4)/(3)` Now, `underset(xto-2)lim(3x^(2)+13x+14)/((x+2)(x-1))=underset(xto-2)lim((3x+7)(x+2))/((x+2)(x-1))=-(1)/(3)`

`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`