Play with LIMITS

$Let,\quad r\in { R }^{ + }\quad be\quad any\quad element.\\ Let\quad us\quad define\quad a\quad function,\quad f:\left[ 0,r \right] \rightarrow \left[ 0,r \right] \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad defined\quad by,\quad f\left( x \right) =\sqrt { { r }^{ 2 }-{ x }^{ 2 } } \\ now,\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \underset { n\longrightarrow \infty }{ lim } { a }_{ n }=\quad \underset { n\longrightarrow \infty }{ lim } \sum _{ k=1 }^{ n }{ \cfrac { 2 }{ n } \left( \sqrt { \left[ 1+{ \left\{ \sqrt { { n }^{ 2 }-\left( k-1 \right) ^{ 2 } } -\sqrt { { n }^{ 2 }-{ k }^{ 2 } } \right\} }^{ 2 } \right] } \right) } \\ \quad \quad \quad \quad \quad \quad =\quad \frac { 4\underset { n\longrightarrow \infty }{ lim } \sum _{ k=1 }^{ n }{ \left( \sqrt { \left[ { \left( \frac { k-1 }{ n } r-\frac { k }{ n } r \right) }^{ 2 }+{ \left\{ f\left( \frac { k-1 }{ n } r \right) -f\left( \frac { k }{ n } r \right) \right\} }^{ 2 } \right] } \right) } }{ 2r } \\ \quad \quad \quad \quad \quad \quad =\quad \frac { 2\pi r }{ 2r } \\ \quad \quad \quad \quad \quad \quad =\quad \pi \\ Hence,\quad \left\lfloor \underset { n\longrightarrow \infty }{ lim } { a }_{ n } \right\rfloor =\left\lfloor \pi \right\rfloor =\left\lfloor 3.14159265359... \right\rfloor =3$