# limxrarra (x^2-a^2)/(sqrtx-sqrta)=32 What value should "a" have?

Sep 4, 2017

The value of $a = 4$

#### Explanation:

To calculate the limit

${\lim}_{x \to a} \frac{{x}^{2} - {a}^{2}}{\sqrt{x} - \sqrt{a}} = \frac{{a}^{2} - {a}^{2}}{\sqrt{a} - \sqrt{s}} = \frac{0}{0}$

Which is an an indeterminate form

We apply l'Hôpital's rule

${\lim}_{x \to a} \frac{{x}^{2} - {a}^{2}}{\sqrt{x} - \sqrt{a}} = {\lim}_{x \to a} \frac{\left({x}^{2} - {a}^{2}\right) '}{\left(\sqrt{x} - \sqrt{a}\right) '}$

$= {\lim}_{x \to a} \frac{2 x}{\frac{1}{2 \sqrt{x}}}$

$= 4 a \sqrt{a}$

Therefore,

$4 a \sqrt{a} = 32$

${a}^{\frac{3}{2}} = \frac{32}{4} = 8$

$a = {8}^{\frac{2}{3}}$

$a = {2}^{2} = 4$

$a = 4$

Sep 4, 2017

$a = 4$

#### Explanation:

We have:

$L = {\lim}_{x \rightarrow a} \frac{{x}^{2} - {a}^{2}}{\sqrt{x} - \sqrt{a}} = 32$

Which we can write as:

$L = {\lim}_{x \rightarrow a} \frac{{x}^{2} - {a}^{2}}{\sqrt{x} - \sqrt{a}} \cdot \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}$
$\setminus \setminus \setminus = {\lim}_{x \rightarrow a} \frac{\left(x + a\right) \left(x - a\right)}{x - a} \cdot \left(\sqrt{x} + \sqrt{a}\right)$
$\setminus \setminus \setminus = {\lim}_{x \rightarrow a} \left(x + a\right) \left(\sqrt{x} + \sqrt{a}\right)$
$\setminus \setminus \setminus = \left(a + a\right) \left(\sqrt{a} + \sqrt{a}\right)$
$\setminus \setminus \setminus = \left(2 a\right) \left(2 \sqrt{a}\right)$
$\setminus \setminus \setminus = 4 {a}^{\frac{3}{2}}$

So if:

$L = 32 \implies 4 {a}^{\frac{3}{2}} = 32$
$\therefore {a}^{\frac{3}{2}} = 8$
$\therefore a \setminus \setminus = 4$

Sep 8, 2017

$a = 4$

#### Explanation:

Factor as a difference of squares multiple times. You may want to recall that ${u}^{2} - {v}^{2} = \left(u + v\right) \left(u - v\right)$. Expand the right hand side to confirm this equality. Using this identity:

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

We'll again use ${u}^{2} - {v}^{2} = \left(u + v\right) \left(u - v\right)$, but here $u = \sqrt{x}$ and $v = \sqrt{a}$.

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

$= {\lim}_{x \rightarrow a} \frac{\left(x + a\right) \left(\sqrt{x} + \sqrt{a}\right) \left(\sqrt{x} - \sqrt{a}\right)}{\sqrt{x} - \sqrt{a}}$

Now the $\sqrt{x} - \sqrt{a}$ terms cancel one another out:

$= {\lim}_{x \rightarrow a} \left(x + a\right) \left(\sqrt{x} + \sqrt{a}\right)$

$= {\lim}_{x \rightarrow a} \left(a + a\right) \left(\sqrt{a} + \sqrt{a}\right)$

$= \left(2 a\right) \left(2 \sqrt{a}\right)$

$= 4 {a}^{\frac{3}{2}} = 32$

Thus:

${a}^{\frac{3}{2}} = 8$

$a = {\left({2}^{3}\right)}^{\frac{2}{3}} = {2}^{2} = 4$