माना कि `x!1` के लिए `f(x)=(1-x(1+|1-x|))/(|1-x|)cos(1/(1-x))`
तब

Options

A. `lim_(x to 1^(+)) f(x)=0`
B. `lim_(x to 1^(-)) f(x)` का अस्तित्व नहीं है
C. `lim_(x to 1^(-)) f(x)=0`
D. `lim_(x to 1^(+)) f(x)` का अस्तित्व नहीं है

Explanation

`f(x)=(1-x(1+|1-x|))/(|1-x|) cos ((1)/(1-x))`
अब `underset(x to 1^(-))lim f(x) =underset(x to 1^(-))lim (1-x(1+1-x))/(1-x) cos ""((1)/(1-x))`
`=underset(x to 1^(-))lim (1-x) cos ((1)/(1-x))=0`
`underset(x to 1^(+)) lim f(x)=underset(x to 1^(+)) (1-x(1-1+x))/(x-1) cos ""((1)/(1-x))`
`=underset(x to 1^(+))lim -(x+1).cos ((1)/(x+1))` जोकि विधमान नहीं है