Question

Evaluate

$\begin{gathered} \rm \lim_{x \to \sqrt{10} } \: \frac{ \sqrt{7 - 2x} - ( \sqrt{5} - \sqrt{2} )}{ {x}^{2} - 10} \\ \end{gathered}\ \textless \ br /\ \textgreater$
​

Answer 1

$\Huge\text{ -\dfrac{5\sqrt{2}+2\sqrt{5}}{60} }$

$\Large\text{\underline{\underline{Question}}}$

Find out the limiting value: -

$\text{ \cdots\longrightarrow\displaystyle\lim_{x\to\sqrt{10}}\dfrac{\sqrt{7-2x}-(\sqrt{5}-\sqrt{2})}{x^{2}-10} }$.

$\Large\text{\underline{\underline{Explanation}}}$

Let's first see if this irrational expression is continuous.

We can see that -

$\cdots\longrightarrow7-2\sqrt{10}=(\sqrt{5}-\sqrt{2})^{2}.$

So, substituting $x=\sqrt{10}$ results in $\dfrac{0}{0}$, which is an indeterminate form and discontinuous.

There exists a solution to radical expressions.

We solve indeterminate forms of radicals in the following technique: -

$\text{ \cdots\longrightarrow\boxed{f(x)-g(x)=\dfrac{\{f(x)\}^{2}-\{g(x)\}^{2}}{f(x)+g(x)}.} }$

So, let's fix this indeterminate form.

$\text{ \cdots\longrightarrow\boxed{\begin{aligned}&\dfrac{\sqrt{7-2x}-(\sqrt{5}-\sqrt{2})}{x^{2}-10}\\\\&=\dfrac{\sqrt{7-2x}-(\sqrt{5}-\sqrt{2})}{x^{2}-10}\times\dfrac{\sqrt{7-2x}+(\sqrt{5}-\sqrt{2})}{\sqrt{7-2x}+(\sqrt{5}-\sqrt{2})}\\\\&=\dfrac{(7-2x)-(\sqrt{5}-\sqrt{2})^{2}}{(x^{2}-10)\{\sqrt{7-2x}+(\sqrt{5}-\sqrt{2})\}}.\end{aligned}} }$

On further simplifying, -

$\text{ \cdots\longrightarrow\boxed{\begin{aligned}&=\dfrac{(7-2x)-(7-2\sqrt{10})}{(x^{2}-10)\{\sqrt{7-2x}+(\sqrt{5}-\sqrt{2})\}}\\\\&=\dfrac{-2x+2\sqrt{10}}{(x^{2}-10)\{\sqrt{7-2x}+(\sqrt{5}-\sqrt{2})\}}\\\\&=\dfrac{-2(x-\sqrt{10})}{(x+\sqrt{10})(x-\sqrt{10})\{\sqrt{7-2x}+(\sqrt{5}-\sqrt{2})\}}\\\\&=-\dfrac{2}{(x+\sqrt{10})\{\sqrt{7-2x}+(\sqrt{5}-\sqrt{2})\}}.\end{aligned}} }$

Now, we can evaluate the limit as, -

$\text{ \cdots\longrightarrow\boxed{\begin{aligned}\displaystyle&\lim_{x\to\sqrt{10}}-\dfrac{2}{(x+\sqrt{10})\{\sqrt{7-2x}+(\sqrt{5}-\sqrt{2})\}}.\\\\&=-\dfrac{2}{2\sqrt{10}\times\{\sqrt{7-2\sqrt{10}}+(\sqrt{5}-\sqrt{2})\}}\\\\&=-\dfrac{1}{\sqrt{10}\times2\times(\sqrt{5}-\sqrt{2})}\\\\&=-\dfrac{\sqrt{10}(\sqrt{5}+\sqrt{2})}{60}\\\\&=-\dfrac{5\sqrt{2}+2\sqrt{5}}{60}.\end{aligned}} }$

So,

$\text{ \cdots\longrightarrow\boxed{\begin{aligned}\displaystyle&\lim_{x\to\sqrt{10}}\dfrac{\sqrt{7-2x}-(\sqrt{5}-\sqrt{2})}{x^{2}-10}=-\dfrac{5\sqrt{2}+2\sqrt{5}}{60}.\end{aligned}} }$

Answer 2

Answer:

This is the answer of your question :-