The correct option is D lim_x→ 2f(x)= 1 At x= 1 lim _ x→ 1 ^ f( x ) #160; =lim _ x→ 1 ^ ( 1) = 1 lim _ x→ 1 ^ + f( x ) #160 ...

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Substitution Method to Remove Indeterminate Form

If fx=-1, ...

Question

If $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} - 1, & x < - 1 x^{3}, & - 1 \leq x \leq 1 1 - x, & 1 < x < 2 3 - x^{2} & x > 2 \end{matrix}$ , then

lim x \to - 1 f (x) = 1

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lim x \to 1 f (x) = 1

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lim x \to 2 f (x) = - 1

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lim x \to 2^{-} f (x) = 0

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Solution

The correct option is D $lim x \to 2 f (x) = - 1$
At $x = - 1$
$lim x \to {- 1}^{-} f (x) = lim x \to {- 1}^{-} (- 1) = - 1$
$lim x \to {- 1}^{+} f (x) = lim x \to {- 1}^{+} (x^{3}) = {(- 1)}^{3} = - 1$
$\Rightarrow lim x \to - 1 = - 1$
At $x = 1$
$lim x \to 1^{-} f (x) = lim x \to 1^{-} x^{3} = 1^{3} = 1$
$lim x \to 1^{+} f (x) = lim x \to 1^{+} (1 - x) = (1 - 1) = 0$
LHL $\neq$ RHL
$\Rightarrow lim x \to 1 f (x)$ does not exist
At $x = 2$
$lim x \to 2^{-} f (x) = lim x \to 2^{-} (1 - x) = 1 - 2 = - 1$
$lim x \to 2^{+} f (x) = lim x \to 2^{+} (3 - x^{2}) = 3 - 4 = - 1$
$\Rightarrow lim x \to 2 f (x) = - 1$

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