Bài tập 29 trang 159 SGK Toán 11 NC
Bài tập 29 trang 159 SGK Toán 11 NC
Cho hàm số
\(f\left( x \right) = \left\{ \begin{array}{l}
2\left| x \right| - 1\,\,\,\,\,\,\,\,khi\,\,x \le - 2\\
\sqrt {2{x^2} + 1} \,\,\,\,khi\,\,x > - 2
\end{array} \right.\)
Tìm \(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right),\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right)\) và \(\mathop {\lim }\limits_{x \to - 2} f\left( x \right)\) (nếu có).
Ta có:
\(\begin{array}{*{20}{l}}
\begin{array}{l}
\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} \left( {2\left| x \right| - 1} \right)\\
= 2.\left| { - 2} \right| - 1 = 3
\end{array}\\
{\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \sqrt {2{x^2} + 1} = 3}\\
{ \Rightarrow \mathop {\lim }\limits_{x \to - 2} f\left( x \right) = 3}
\end{array}\)
-- Mod Toán 11
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