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Indeterminate Forms

Evaluate the ...

Question

Evaluate the limit:

$lim x \to 1 \frac{x^{3} + 3 x^{2} - 6 x + 2}{x^{3} + 3 x^{2} - 3 x - 1}$

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Solution

Given: $lim x \to 1 \frac{x^{3} + 3 x^{2} - 6 x + 2}{x^{3} + 3 x^{2} - 3 x - 1}$ becomes $\frac{0}{0}$ form

Now numerator
$\Rightarrow x^{3} + 3 x^{2} + 6 x + 2 = (x - 1) (x^{2} + 4 x - 2)$

Also, denominator

$\Rightarrow x^{3} + 3 x^{2} - 3 x - 1 = (x - 1) (x^{2} + 4 x + 1)$

$∴ lim x \to 1 \frac{x^{3} + 3 x^{2} - 6 x + 2}{x^{3} + 3 x^{2} - 3 x - 1}$

$= lim x \to 1 \frac{(x - 1) (x^{2} + 4 x - 2)}{(x - 1) (x^{2} + 4 x + 1)}$

$\Rightarrow lim x \to 1 \frac{(x^{2} + 4 x - 2)}{(x^{2} + 4 x + 1)}$

= $\frac{(1^{2} + 4 (1) - 1)}{(1^{2} + 4 (1) + 1)} = \frac{3}{6}$

= $\frac{1}{2}$

$T h e r e f o r e,$

$lim x \to 1 \frac{x^{3} + 3 x^{2} - 6 x + 2}{x^{3} + 3 x^{2} - 3 x - 1} = \frac{1}{2}$

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Indeterminate Forms

Standard XIII Mathematics

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