We have, f (x) = |x| - 5 Now, $\lim\limits_{x\to 5^{-}}$ f (x) = $\lim\limits_{x\to 5^{-}}$ (|x| - 5) = $\lim\limits_{x\to 5^{-}}$ (x - 5) [Since |x| = x , for x > 0] = 5 - 5 = 0 and $\lim\limits_{x\to 5^{+}}$ f (x) = $\lim\limits_{x\to 5^{+}}$ (|x| - 5) $\lim\limits_{x\to 5^{+}}$ (x - 5) = 5 - 5 = 0 [Since |x| = x, for x > 0] ∴ $\lim\limits_{x\to 5^{-}}$ f (x) = $\lim\limits_{x\to 5^{+}}$ f (x) Thus $\lim\limits_{x\to 5}$ f (x) = 0.