1.

If `f(x)={{:(,x([(1)/(x)]+[(2)/(x)]+.....+[(n)/(x)]),x ne 0),(,k,x=0):}` and `n in N`. Then the value of k for which f(x) is continuous at x=0 isA. nB. n+1C. `n(n+1)`D. `(n(n+1))/(2)

Answer» Correct Answer - D
For any `x gt 0`, we know that
`(1)/(x)-1 lt [(1)/(x)] le (1)/(x) Rightarrow 1-x lt x[(1)/(x)] le 1`
`(2)/(x)-1 lt [(2)/(x)] le (2)/(x) Rightarrow 2-x le x[(2)/(x)] le 2`
`(n)/(x)-1 lt [(n)/(x)] le (n)/(x) Rightarrow n-x lt x[(n)/(x)] le n`
`therefore (n(n+1))/(2)-nx lt x ([(1)/(x)]+[(2)/(x)]+....+[(n)/(x)])le (n(n+1))/(2)`
But
`underset(x to 0)lim {(n(n+1))/(2)-nx}=(n(n+1))/(2)and, underset(x to )lim (n(n+1))/(2)=(n(n+1))/(2)`
Therefore, by Sandwhich theoram, we obtain
`underset(x to 0)lim x([(1)/(x)]+[(2)/(x)]+.....+[(n)/(x)])=(n(n+1))/(2)`


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