# Question ff93d

May 12, 2017

$4$

#### Explanation:

Multiply the numerator and denominator by the conjugate of the denominator:

$= {\lim}_{x \rightarrow 2} \frac{\left(x - \sqrt{8 - {x}^{2}}\right) \left(\sqrt{{x}^{2} + 12} + 4\right)}{\left(\sqrt{{x}^{2} + 12} - 4\right) \left(\sqrt{{x}^{2} + 12} + 4\right)}$

$= {\lim}_{x \rightarrow 2} \frac{\left(x - \sqrt{8 - {x}^{2}}\right) \left(\sqrt{{x}^{2} + 12} + 4\right)}{\left({x}^{2} + 12\right) - 16}$

$= {\lim}_{x \rightarrow 2} \frac{\left(x - \sqrt{8 - {x}^{2}}\right) \left(\sqrt{{x}^{2} + 12} + 4\right)}{{x}^{2} - 4}$

The denominator is still $0$ as $x \rightarrow 2$, so let's also try multiplying by the conjugate of the original numerator:

=lim_(xrarr2)((x-sqrt(8-x^2))(x+sqrt(8-x^2))(sqrt(x^2+12)+4))/((x^2-4)(x+sqrt(8-x^2))

=lim_(xrarr2)((x^2-(8-x^2))(sqrt(x^2+12)+4))/((x^2-4)(x+sqrt(8-x^2))

=lim_(xrarr2)((2x^2-8)(sqrt(x^2+12)+4))/((x^2-4)(x+sqrt(8-x^2))

=lim_(xrarr2)(2(x^2-4)(sqrt(x^2+12)+4))/((x^2-4)(x+sqrt(8-x^2))

=lim_(xrarr2)(2(sqrt(x^2+12)+4))/(x+sqrt(8-x^2)#

Now we can evaluate the limit:

$= \frac{2 \left(\sqrt{4 + 12} + 4\right)}{2 + \sqrt{8 - 4}}$

$= \frac{2 \left(4 + 4\right)}{2 + 2}$

$= 4$

May 12, 2017

${\lim}_{x \to 2} \frac{x - \sqrt{8 - {x}^{2}}}{\sqrt{{x}^{2} + 12} - 4}$

$= {\lim}_{x \to 2} \frac{x - \sqrt{8 - {x}^{2}}}{\sqrt{{x}^{2} + 12} - 4} \times \frac{x + \sqrt{8 - {x}^{2}}}{\sqrt{{x}^{2} + 12} + 4} \times \frac{\sqrt{{x}^{2} + 12} + 4}{x + \sqrt{8 - {x}^{2}}}$

$= {\lim}_{x \to 2} \frac{{x}^{2} - 8 + {x}^{2}}{{x}^{2} + 12 - {4}^{2}} \times \frac{\sqrt{{x}^{2} + 12} + 4}{x + \sqrt{8 - {x}^{2}}}$

$= {\lim}_{x \to 2} \frac{2 \left(\cancel{{x}^{2} - 4}\right)}{\left(\cancel{{x}^{2} - 4}\right)} \times \frac{\sqrt{{x}^{2} + 12} + 4}{x + \sqrt{8 - {x}^{2}}}$

[$\text{ "as" } x \to 2 \implies x \ne 2 \implies {x}^{2} - 4 \ne 0$ ]

so the limit becomes

$= {\lim}_{x \to 2} \frac{2 \left(\sqrt{{x}^{2} + 12} + 4\right)}{x + \sqrt{8 - {x}^{2}}}$

$= \frac{2 \left(\sqrt{{2}^{2} + 12} + 4\right)}{2 + \sqrt{8 - {2}^{2}}}$

$= \frac{2 \left(\sqrt{16} + 4\right)}{2 + \sqrt{8 - 4}}$

$= \frac{2 \times 8}{2 + 2} = 4$