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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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  • 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 by thanvigandhi_1
edited Mar 26, 2013 by thanvigandhi_1
 

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