If F(x) = `{(sin{cosx})/(x-pi/2),x!=pi/2 and 1,x=pi/2,` where {.} represents the fractional part function, then `lim_(xto pi//2)f(x)` is

Options

A. continuous at `x=pi//2`
B. `underset(x to pi//2)lim f(x)` but f(x) is not continuous at `x=pi//2`
C. `underset(x to pi//2)lim f(x)` does not exist
D. `underset(x to pi//2^(-))lim f(x) =-1`

Answer

B

Explanation

We have
`underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim f(pi//2-h)`
`Rightarrow underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim (sin {cos (pi//2-h)})/(-h)`
`Rightarrow underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim (sin{sin h})/(-h)2`
`Rightarrow underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim (sin(sin h))/(-h)`
`Rightarrow underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim (sin(sin h))/(sin h)xx(sin h)/(h)=-1` and
`underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim f(pi//2+h)`
`Rightarrow underset(xto pi//2^(+))lim f(x)=underset(h to 0)lim (sin {cos (pi//2+h)})/((pi)/(2)+h-(pi)/(2))`
`Rightarrow underset(xto pi//2^(+))lim f(x)=underset(h to 0)lim (sin{-sin h})/(h)`
`Rightarrow underset(xto pi//2^(+))lim f(x)=underset(h to 0)lim (sin{sin h})/(h)xx(sin h)/(h)=-1`
`"Clearly", underset(x to pi//2^(-))lim f(x)=underset(x to pi//2^(+))lim f(x) ne f((pi)/(2))`
`"So", underset(x to pi//2)lim` f(x) exists but f(x) is not continuous at `x=pi//2`.