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Preeti Dabral 3 years ago (2983787)
{tex}\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} 2x + 1{/tex}
{tex}\mathop {\lim }\limits_{h \to 0} \left[ {2\left( {2 - h} \right) + 1} \right]{/tex} = 5
{tex}= \mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {3x -1} \right){/tex}
{tex} = \mathop {\lim }\limits_{h \to 0} 3\left( {2 + h} \right) - 1{/tex} = 5
In given that question f(x) is continuous at x=2,therefore
{tex} \mathop {\lim }\limits_{x \to {2^ - }} f(x) = f(2) = \mathop {\lim }\limits_{x \to {2^ + }} f(x){/tex}
5 = k
{tex}\Rightarrow k = 5{/tex}
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Karishma Das 3 years, 2 months ago (13388375)
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