Question

$lim \: x \: \to \: 2 \: \frac{x + {x}^{2} + {x}^{3} - 14}{ {x}^{2} - 4 }$

Please don't use method of differentiation.

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Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 2} \: \frac{x + {x}^{2} + {x}^{3} - 14}{ {x}^{2} - 4}$

If we substitute directly x = 2, we get

$\rm \: = \: \dfrac{2 + {2}^{2} + {2}^{3} - 14}{ {2}^{2} - 4}$

$\rm \: = \:\dfrac{2 + 4 + 8 - 14}{4 - 4}$

$\rm \: = \:\dfrac{14 - 14}{4 - 4}$

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

which is indeterminant form.

Consider, again

$\rm :\longmapsto\:\displaystyle\lim_{x \to 2} \: \frac{x + {x}^{2} + {x}^{3} - 14}{ {x}^{2} - 4}$

can be rewritten as

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

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

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

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

We know,

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

and

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

So, using these, we get

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

On cancelation the factor x - 2 from numerator and denominator, we get

$\rm \: = \:\displaystyle\lim_{x \to 2} \frac{1 + (x + 2) + ( {x}^{2} + 2x + 4)}{(x + 2)}$

$\rm \: = \:\dfrac{1 + (2 + 2) + (4 + 4 + 4)}{2 + 2}$

$\rm \: = \:\dfrac{17}{4}$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 2} \: \frac{x + {x}^{2} + {x}^{3} - 14}{ {x}^{2} - 4} = \frac{17}{4}}}$

Additional Information :-

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

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

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{log(1 + x)}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{ {e}^{x} - 1}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1}{x} = loga \: }}$