If the following functions are continuous at x=1 then ascending order of a,b,c 1) f(x)=x3−1x−1 for x≠1, f(1)=a 2) f(x)=x−2−1x−1 for x≠1, f(1)=b 3) f(x)=x−1x2−1 for x≠1, f(1)=c

a=Ltx→1x3−1x−1=Ltx→1(x2+x+1)=1+1+1=3 b=Ltx→1x−2−1x−1=Ltx→11-x2x2(x−1)=Ltx→1-(x+1)x2=-2c=Ltx→1x−1x2−1=Ltx→1(x−1)(x−1)(x+1)=12

If the following functions are continuous at x=1 then ascending order of a,b,c 1) f(x)=x3−1x−1 for x≠1, f(1)=a 2) f(x)=x−2−1x−1 for x≠1, f(1)=b 3) f(x)=x−1x2−1 for x≠1, f(1)=c

a=Ltx→1x3−1x−1=Ltx→1(x2+x+1)=1+1+1=3 b=Ltx→1x−2−1x−1=Ltx→11-x2x2(x−1)=Ltx→1-(x+1)x2=-2c=Ltx→1x−1x2−1=Ltx→1(x−1)(x−1)(x+1)=12

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If the following functions are continuous at $x = 1$ then ascending order of $a, b, c$
1) $f (x) = \frac{x^{3} - 1}{x - 1}$ for $x \neq 1, f (1) = a$

2) $f (x) = \frac{x^{- 2} - 1}{x - 1}$ for $x \neq 1, f (1) = b$

3) $f (x) = \frac{x - 1}{x^{2} - 1}$ for $x \neq 1, f (1) = c$

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a, b, c

b, c, a

a, c, b

c, a, b

answer is B.

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Detailed Solution

$a = \underset{x \to 1}{L t} \frac{x^{3} - 1}{x - 1} = \underset{x \to 1}{L t} (x^{2} + x + 1) = 1 + 1 + 1 = 3$

$b = \underset{x \to 1}{L t} \frac{x^{- 2} - 1}{x - 1} = \underset{x \to 1}{L t} \frac{1 - x^{2}}{x^{2} (x - 1)} = \underset{x \to 1}{L t} \frac{- (x + 1)}{x^{2}} = - 2$

$c = \underset{x \to 1}{L t} \frac{x - 1}{x^{2} - 1} = \underset{x \to 1}{L t} \frac{(x - 1)}{(x - 1) (x + 1)} = \frac{1}{2}$

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If the following functions are continuous at x=1 then ascending order of a,b,c 1) f(x)=x3−1x−1 for x≠1, f(1)=a 2) f(x)=x−2−1x−1 for x≠1, f(1)=b 3) f(x)=x−1x2−1 for x≠1, f(1)=c

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If the following functions are continuous at $x = 1$ then ascending order of $a, b, c$
1) $f (x) = \frac{x^{3} - 1}{x - 1}$ for $x \neq 1, f (1) = a$

2) $f (x) = \frac{x^{- 2} - 1}{x - 1}$ for $x \neq 1, f (1) = b$

3) $f (x) = \frac{x - 1}{x^{2} - 1}$ for $x \neq 1, f (1) = c$