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.
Câu 392281: 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.
A. \(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = 3\)
\(\mathop {\lim }\limits_{x \to \left( { - 2} \right)} f\left( x \right) = 3\).
B. \(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = 2\)
\(\mathop {\lim }\limits_{x \to \left( { - 2} \right)} f\left( x \right) = 2\).
C.
\(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = 0\)
\(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = -1\)
D. \(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = + \infty\)
\(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = -1\)
- Tính lần lượt \(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right),\,\,\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right)\).
- Nếu \(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right)\) thì tồn tại \(\mathop {\lim }\limits_{x \to \left( { - 2} \right)} f\left( x \right)\) và \(\mathop {\lim }\limits_{x \to \left( { - 2} \right)} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right)\).
- Nếu \(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right)\) thì không tồn tại \(\mathop {\lim }\limits_{x \to \left( { - 2} \right)} f\left( x \right)\).
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Đáp án : A(0) bình luận (0) lời giải
Giải chi tiết:
Ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} \left( {2\left| x \right| - 1} \right) = 2\left| { - 2} \right| - 1 = 3\\\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \sqrt {2{x^2} + 1} = \sqrt {2.{{\left( { - 2} \right)}^2} + 1} = 3\end{array}\)
Vì \(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = 3\) nên \(\mathop {\lim }\limits_{x \to \left( { - 2} \right)} f\left( x \right) = 3\).
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