Therefore, $\mathop {\lim }\limits_{x \to {0^ + }} \,f(x) = 0 = \mathop {\lim }\limits_{x \to {0^ - }} \,f(x) = f(0)$
Hence $f(x)$ is continuous at $x = 0.$
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If
$\text{y}=\text{e}^{-\text{x}}(\text{A}\cos\text{x}+\text{B}\sin\text{x}),$ then y is a solution of:$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}=0$
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\text{y}=0$