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For what value of \( \lambda \) is the function defined by \(f\) defined by $ f(x) = \left\{ \begin{array} {1 1} \lambda(x^2 - 2x) ,& \quad\text{ if $ x $ \(\leq 0\)}\\ 4x + 1,& \quad \text{if $x$ > 0}\\ \end{array} \right.$ What about continuity at \( x = 1\)?

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  • If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
At $x=1$
$\lim\limits_{\large x\to 1}f(x)=\lim\limits_{\large x\to 1}(4x+1)$
$\quad\quad\quad\;\;=4+1$
$\quad\quad\quad\;\;=5$
$\Rightarrow f$ is not continuous at $x=0$ for any value of $\lambda$ but continuous at $x=1$ for all values of $\lambda$.
answered Aug 19, 2013 by sreemathi.v
 

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