Let `f(x)={((tan^(2){x})/(x^(2)-[x]^(2)),"for"xgt0),(1/(sqrt({x}cot{x})),"for"xlt0):}` where `[x]` is the step up function and `{x}` is the fractional part function of x then

Options

A. `lim_(xto0^+)f(x)=1`
B. `lim_(xto0^-)f(x)=1`
C. `cot^-1{lim_(xto0^-)f(x)}^2=(pi)/(4)`
D. All of these

Answer

D

Explanation

We have,
`f(x) =(tan^2{x})/(x^2-[x]^2) " for " xgt 0`
`lim_(xto0^+) f(x)=lim_(xto0^+)(tan^2{x})/(x^2-[x]^2)=lim_(xto0^+)(tan^2x)/(x^2)=1`.
So, option (a) is correct
`{lim_(xto0^-)f(x)}^2=1^2=1`.
`therefore cot^-1{lim_(xto0^-)f(x)}^2=cot^-1(1)=(pi)/(4)`
So, option c is also correct.