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$f(x)=\left\{\begin{array}{1 1}\large\frac{x^5-32}{x-2}&x\neq 2\\k&x=2\end{array}\right.$ is continuous at $x=2$ then $k=$

$(a)\;16\qquad(b)\;80\qquad(c)\;32\qquad(d)\;8$

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$\lim\limits_{x\to 2}f(x)=\lim\limits_{x\to 2}\large\frac{x^5-32}{x-2}$
$\Rightarrow \lim\limits_{x\to 2}\large\frac{x^5-2^5}{x-2}$
$\Rightarrow 5(2)^4$
$\Rightarrow 80$
Since $f(x)$ is continuous at $x=2$
$\lim\limits_{x\to 2}f(x)=f(2)$
$k=80$
Hence (b) is the correct answer.
answered Dec 30, 2013 by sreemathi.v
 

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