Official solution
\[f(x) = [x] - \left[ {{x \over 4}} \right]\]
Here check continuty at x = 4
LHL = \[\mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right)\]
= \[\mathop {\lim }\limits_{x \to {4^ - }} \left[ x \right] - \left[ {{x \over 4}} \right]\]
= \[\mathop {\lim }\limits_{h \to 0} \left[ {4 - h} \right] - \left[ {{{4 - h} \over 4}} \right]\]
= 3 - 0 = 3
h \[ \to \] 0 means h > 0 (very small positive value)
So 4 - h = 3.something (means less than 4)
\[ \therefore \] [4 - h] = [3.something] = 3
and \[\left[ {{{4 - h} \over 4}} \right]\] = [0.something] = 0
RHL = \[\mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right)\]
= \[\mathop {\lim }\limits_{x \to {4^ + }} \left[ x \right] - \left[ {{4 \over x}} \right]\]
= \[\mathop {\lim }\limits_{h \to 0} \left[ {4 + h} \right] - \left[ {{{4 + h} \over 4}} \right]\]
= 4 - 1 = 3
Here [4 + h] = [4.something] = 4
and \[\left[ {{{4 + h} \over 4}} \right]\] = [1.something] = 1
And f(4) = \[\left[ 4 \right] - \left[ {{4 \over 4}} \right]\] = 4 - 1 = 3
As LHL = RHL = f(4)
\[ \therefore \] f is continuous at x = 4.