Question

Lim -> 2 x^2-6x+8/x^3-8

Please answer this question

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 2} \frac{ {x}^{2} - 6x + 8}{ {x}^{3} - 8}$

If we substitute directly x = 2, we get

$\rm \: = \:\dfrac{ {2}^{2} - 2 \times 6 + 8}{ {2}^{3} - 8}$

$\rm \: = \:\dfrac{4 - 12 + 8}{8 - 8}$

$\rm \: = \:\dfrac{0}{0}$

which is indeterminant form.

So, to evaluate this

$\rm :\longmapsto\:\displaystyle\lim_{x \to 2} \frac{ {x}^{2} - 6x + 8}{ {x}^{3} - 8}$

We use Method of Factorization.

$\rm \: = \:\displaystyle\lim_{x \to 2} \frac{ {x}^{2} - 4x - 2x + 8 }{ {x}^{3} - {2}^{3} }$

We know,

$\boxed{ \tt{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}) \: }}$

So, using this identity, we get

$\rm \: = \:\dfrac{x(x - 4) - 2(x - 4)}{(x - 2)( {x}^{2} + 2x + 4)}$

$\rm \: =\displaystyle\lim_{x \to 2} \:\dfrac{(x - 4) \: \cancel{(x - 2)}}{ \cancel{(x - 2)} \: \: ( {x}^{2} + 2x + 4)}$

$\rm \: = \:\displaystyle\lim_{x \to 2} \frac{x - 4}{ {x}^{2} + 2x + 4 }$

$\rm \: = \:\dfrac{2 - 4}{4 + 4 + 4}$

$\rm \: = \:\dfrac{ - 2}{4 + 4 + 4}$

$\rm \: = \:\dfrac{ - 2}{12}$

$\rm \: = \:\dfrac{ - 1}{6}$

Thus,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 2} \frac{ {x}^{2} - 6x + 8}{ {x}^{3} - 8} = - \frac{1}{6} \: }}}$

More to know

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$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{tanx}{x} = 1 \: }}$

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