Tìm a để hàm số \(f\left( x \right) = \left\{ \begin{array}{l}{x^2} + ax\;\;khi\;x > 3\\3{x^2} + 1\;\;\;khi\;x \le 3\end{array} \right.\) có giới hạn khi \(x \to 3\).
\(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = L\) khi và chỉ khi \(\mathop {\lim }\limits_{x \to x_0^ + } f\left( x \right) = \mathop {\lim }\limits_{x \to x_0^ - } f\left( x \right) = L\)
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Ta có: \(\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} \left( {{x^2} + ax} \right) = 9 + 3a\), \(\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ - }} \left( {3{x^2} + 1} \right) = 28\)
Do đó, hàm số f(x) có giới hạn khi \(x \to 3\) khi \(\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right)\)
Suy ra \(9 + 3a = 28 \Leftrightarrow a = \frac{{19}}{3}\)