If `m, n in I_(0)` and `lim_(xto0) (tan2x-nsinx)/(x^(3))` = some integer, then find the value of n and also the value of limit.

If `m, n in I_(0)` and `lim_(xto0) (tan2x-nsinx)/(x^(3))` = some integ

1.	If `m, n in I_(0)` and `lim_(xto0) (tan2x-nsinx)/(x^(3))` = some integer, then find the value of n and also the value of limit.
Answer» `L=underset(xto0)lim(tan2x-nsinx)/(x^(3))` `underset(xto0)lim(sin2x-nsinxcos2x)/(x^(3)cos2x)` `=underset(xto0)lim("sin"x)/(x)((2cosx-ncos2x))/(x^(2))=(1)/(cos2x)` `=underset(xto0)lim((2cosx-ncos2x))/(x^(2))` Now, for `xto0`,`x^(2)to0`. Therefore, for `xto0, 2cosx-ncos2xto0.` So, n=2. For, n=2. `L=underset(xto0)lim((2cosx-ncos2x))/(x^(2))` `=4underset(xto0)lim("sin"(x)/(2)"sin"(3x)/(2))/(x^(2))` =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`
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`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`