Prove that `lim_(xto2) [x]` does not exists, where [.] represents the greatest integer function.

Prove that `lim_(xto2) [x]` does not exists, where [.] represents the

1.	Prove that `lim_(xto2) [x]` does not exists, where [.] represents the greatest integer function.
Answer» We know that greatest integer function changes its value at integral values of x. So, let us find LHL and RHL. `LHL=underset(xto2)lim[x]=underset(htoo)lim[2-h]` Sine h is positive and infinitely small `(2-h)` is value very close to 2 but smaller than 2 (say 1.99999) `:." "[2-h]=1,` when h approaches to zero. Thus, LHL=1 `RHL=underset(xto2)lim[x]=underset(hto0)lim[2+h]=[2.00001]=2` We observe that `LHLneRHL.` So, `underset(xto2)lim[x]` does not exist.

`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`