Question -16

Checking continuity at x=1:

Left Hand Limit: $\lim _{ x\rightarrow { 1 }^{ - } }{ \left\lfloor x \right\rfloor } +\sqrt { \left\{ x \right\} } \quad (\because \quad x-\left\lfloor x \right\rfloor =\left\{ x \right\} )$

$=\lim _{ h\rightarrow 0 }{ \left\lfloor 1-h \right\rfloor } +\sqrt { \left\{ 1-h \right\} } \\ \\ =0+\lim _{ h\rightarrow 0 }{ \sqrt { 1-h } } \\ \\ =1$

Similarly, Right Hand Limit: $\lim _{ x\rightarrow { 1 }^{ + } }{ \left\lfloor x \right\rfloor } +\sqrt { \left\{ x \right\} } \quad \\ \\ =\lim _{ h\rightarrow 0 }{ \left\lfloor 1+h \right\rfloor } +\sqrt { \left\{ 1+h \right\} } \\ \\ =1+\lim _{ h\rightarrow 0 }{ \sqrt { h } } \\ \\ =1$

Also, value of function at x=1: $f(1)=\left\lfloor 1 \right\rfloor +\sqrt { \left\{ 1 \right\} } =1+0=1$

Since LHL=RHL=f(1), hence f(x) is continuous at x=1.

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Now, checking differentiability at x=1:

Left Hand Derivative: $\lim _{ x\rightarrow { 1 }^{ - } }{ \cfrac { f(x)-f(1) }{ x-1 } } \quad \\ \\ =\lim _{ h\rightarrow 0 }{ \cfrac { f(1-h)-f(1) }{ 1-h-1 } } \\ =\lim _{ h\rightarrow 0 }{ \cfrac { \left\lfloor 1-h \right\rfloor -\sqrt { \left\{ 1-h \right\} } -1 }{ -h } } \\ =\lim _{ h\rightarrow 0 }{ \cfrac { 0-\sqrt { 1-h } -1 }{ -h } } \\ =\lim _{ h\rightarrow 0 }{ \cfrac { -(\sqrt { 1-h } +1) }{ -h } } \\ =\lim _{ h\rightarrow 0 }{ \cfrac { (\sqrt { 1-h } +1) }{ h } } \\$

Right Hand derivative: $\lim _{ x\rightarrow { 1 }^{ + } }{ \cfrac { f(x)-f(1) }{ x-1 } } \quad \\ \\ =\lim _{ h\rightarrow 0 }{ \cfrac { f(1+h)-f(1) }{ 1+h-1 } } \\ =\lim _{ h\rightarrow 0 }{ \cfrac { \left\lfloor 1+h \right\rfloor -\sqrt { \left\{ 1+h \right\} } -1 }{ h } } \\ =\lim _{ h\rightarrow 0 }{ \cfrac { 1-\sqrt { 1+h } -1 }{ h } } \\ =\lim _{ h\rightarrow 0 }{ \cfrac { -(\sqrt { 1-h } ) }{ h } } \\$

Clearly the value of left hand derivative approaches infinity and that of right hand derivative approaches negative infinity. Hence, f(x) is not differentiable at x=1.

Hence, f(x) is continuous but non-differentiable at x=1.