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If $f(x)=\large\frac{x^2-10x+25}{x^2-7x+10}$ for $x\neq 5$ and $f$ is continuous at $x=5$ then $f(5)=$

$(a)\;0\qquad(b)\;5\qquad(c)\;10\qquad(d)\;25$

1 Answer

$f(5)=\lim\limits_{x\to 5}f(x)=\lim\limits_{x\to 5}\large\frac{x^2-10x+25}{x^2-7x+10}$
$\Rightarrow \lim\limits_{x\to 5}\large\frac{(x-5)^2}{(x-2)(x-5)}$
$\Rightarrow \lim\limits_{x\to 5}\large\frac{(x-5)}{(x-2)}$
$\Rightarrow \large\frac{5-5}{5-2}$
$\Rightarrow 0$
Hence (a) is the correct answer.
answered Dec 30, 2013 by sreemathi.v
 

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