Let`f(3)=4 and f'(3)=5`. Then `lim_(xrarr3) [f(x)]` (where [.] denotes the greatest integer function) is

Options

A. 3
B. 4
C. 5
D. non-existent

Answer

D

Explanation

`f'(3)=underset(hrarr0)(lim)(f(3+4)-f(3))/(h)=underset(hrarr0)(lim)(f(3-4)-f(3))/(-h)`
`therefore" "f(3+h)-f(3)=5h+h=epsilon_(1)`
where `epsilon_(1)rarr0` when `h rarr0`
and `f(3-h)-f(3)=-5h+h=epsilon_(2)`
where `epsilon_(2)rarr0` when `hrarr0`
`therefore" "f(3+h)=f(3)+5h+h=epsilon_(1)=4+5h+hepsilon_(1)`
`therefore" "[f(3+h)]=[4+5h+h epsilon_(1)]=4" when " h gt 0 and h " is very small."`
`therefore" "underset(hrarr0)(lim)[f(3+h)]=4`
Similarly, `[f(3-h)]=[4-5h+hepsilon_(2)]=[4-h(5-epsilon_(2)]=3`
When `hgt0` and h is very small implying `epsilon_(2)` is very small.
`therefore" "underset(hrarr0)(lim)[f(3-h)=3 ne 4=underset(hrarr0)(lim)[f(3+h)]`
`therefore" "underset(xrarr3)(lim)[f(x)]` does not exist.