Let `f: R->[0,oo)` be such that `lim_(x->5) f(x)` exists and `lim_(x->5) ((f(x))^2-9)/sqrt(|x-5|)=0` then `lim_(x->5)f(x)=`A. 1B. 2C. 3D. 0

Let `f: R->[0,oo)` be such that `lim_(x->5) f(x)` exists and `lim_(x->

1.	Let `f: R->[0,oo)` be such that `lim_(x->5) f(x)` exists and `lim_(x->5) ((f(x))^2-9)/sqrt(\|x-5\|)=0` then `lim_(x->5)f(x)=`A. 1B. 2C. 3D. 0
Answer» Correct Answer - C `lim_(xto5)((f(x))^2-9)/(sqrt(\|x-5\|))=0` ` rArr lim_(xto5)(f(x))^2-9=0` ` rArr l^2-9=0, "where" lim_(xto5)f(x)=l` `rArr l= +-3` `rArr l=3 [because f(x)ge 0 " for all " x in R ]` `rArr lim_(xto5)f(x)=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`