If f(x)=x2−10x+25x2−7x+10 for x≠5 and f is continuous at x=5 then ƒ(5)=

f(5)=Ltx→5f(x)=Ltx→5x2−10x+25x2−7x+10(00form using l'hospital's rule) =Ltx→52x−102x−7=10−1010−7=0

If f(x)=x2−10x+25x2−7x+10 for x≠5 and f is continuous at x=5 then ƒ(5)=

f(5)=Ltx→5f(x)=Ltx→5x2−10x+25x2−7x+10(00form using l'hospital's rule) =Ltx→52x−102x−7=10−1010−7=0

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

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$f (5) = \underset{x \to 5}{L t} f (x) = \underset{x \to 5}{L t} \frac{x^{2} - 10 x + 25}{x^{2} - 7 x + 10} (\frac{0}{0} f o r m u \sin g l' h o s p i t a l' s r u l e)$

$= \underset{x \to 5}{L t} \frac{2 x - 10}{2 x - 7} = \frac{10 - 10}{10 - 7} = 0$

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If f(x)=x2−10x+25x2−7x+10 for x≠5 and f is continuous at x=5 then ƒ(5)=

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