If `lim_(x->0)(x^n-sinx^n)/(x-sin^n x)` is non-zero finite, then `n` must be equal to 4 (b) 1 (c) 2 (d) 3

Options

A. `1," if "n=m`
B. `0," if "ngtm`
C. `oo," if "nltm`
D. `n//m, " if "nltm`

Answer

B

Explanation

`underset(xto0)lim(x^(n)sin^(n)x)/(x^(n)-sin^(n)x)=underset(xto0)lim(x^(n)(x-(x^(3))/(3!)+(x^(5))/(5!)-...)^(n))/(x^(n)-(x-(x^(3))/(3!)+(x^(5))/(5!)-...)^(n))`
`=underset(xto0)lim((x-(x^(3))/(3!)+(x^(5))/(5!)-...)^(n))/(1-(1-(x^(2))/(3!)+(x^(4))/(5!)-...)^(n))`
`=underset(xto0)lim(x^(n)(1-(x^(2))/(3!)+(x^(4))/(5!)-...)^(n))/(1-(1-(x^(2))/(3!)+(x^(4))/(5!)-...)^(n))`
For `n=2,`
`=underset(xto0)lim(x^(2)(1-(x^(2))/(3!)+(x^(4))/(5!)-...)^(2))/(1-(1-(x^(2))/(3!)+(x^(4))/(5!)-...)^(2))`
`=underset(xto0)lim(x^(2)(1-(x^(2))/(3!)+(x^(4))/(5!)-...)^(2))/((2-(x^(2))/(3!)+(x^(4))/(5!)-...)((x^(2))/(3!)-(x^(4))/(5!)+...))`
`=(1(1-0+...)^(2))/((2-0+0)((1)/(3!)-0+....))`
`=3`