Cho hàm số
\(f\left( x \right) = \left\{ {\matrix{{2\left| x \right| – 1\,\text{ với }\,x \le – 2,} \cr {\sqrt {2{x^2} + 1} \,\text{ với }\,x > – 2.} \cr} } \right.\)
Tìm \(\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ – }} f\left( x \right),\mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} f\left( x \right)\,\text{ và }\,\mathop {\lim }\limits_{x \to – 2} f\left( x \right)\) (nếu có).
Ta có:
\(\eqalign{
& \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ – }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ – }} \left( {2\left| x \right| – 1} \right) = 2\left| { – 2} \right| – 1 = 3 \cr
& \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} = \mathop {\lim }\limits_{x \to {{\left( { – 2} \right)}^ + }} \sqrt {2{x^2} + 1} = 3 \Rightarrow \mathop {\lim }\limits_{x \to – 2} f\left( x \right) = 3. \cr} \)