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# Show that the function $f(x)= \left\{ \begin{array}{1 1} x+\lambda, & \quad x< 1 \\ \lambda x^2+1, & \quad x\geq1 \end{array} \right.$ is continuous function, regardless of the choice of $\lambda \in R$

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Toolbox:
• For continuous function at $x = 1. LHL = RHL = f(1)$
LHL $\lim\limits_{x \to 1}\: x+\lambda=1+\lambda$
LHL $\lim\limits_{x \to 1}\: \lambda x^2+1=\lambda+1$

$f(1) = \lambda+1$
$\Rightarrow$ is continuous regardsless of choice of $\lambda$

answered Mar 11, 2013
edited Mar 26, 2013