Cho hàm số (fleft( x ight) = left{ egin{array}{l}3x + a - 1,,,,,,,,,,khi,,x le 0\frac{{sqrt {1 + 2x}  - 1}}{x},,,

* Hướng dẫn giải

Ta có:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {o^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {o^ + }} \frac{{\sqrt {1 + 2x}  - 1}}{x}\\
 = \mathop {\lim }\limits_{x \to {o^ + }} \frac{{2x}}{{x\left( {\sqrt {1 + 2x}  + 1} \right)}} = \mathop {\lim }\limits_{x \to {o^ + }} \frac{2}{{\sqrt {1 + 2x}  + 1}} = 1
\end{array}\) 

Và \(\mathop {\lim }\limits_{x \to {o^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {o^ - }} \left( {3x + a - 1} \right) = a - 1\) 

Mặt khác: \(f\left( 0 \right) = a - 1\) 

Hàm số liên tục tại \(x=0\)

\(\begin{array}{l}
 \Leftrightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to {o^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {o^ + }} f\left( x \right)\\
 \Leftrightarrow a - 1 = 1 \Leftrightarrow a = 2
\end{array}\)