Evaluate the expression $\underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{\sqrt{7+2x}-\left(
\sqrt{5}+\sqrt{2} \right)}{{{x}^{2}}-10}$

Hint: You could use either of the two here, you could use L’ Hopital’s rule after confirming this limit has
an indeterminate form, or you could directly work by rationalising the numerator, to do away with the
radicals, and then simply substituting $x$ with its limiting value in the expression you get thereafter.
We’ll use the simple way of rationalising to find out the limit over here, since it’s easier.
Now the given equation is;
$\underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{\sqrt{7+2x}-\left( \sqrt{5}+\sqrt{2} \right)}{{{x}^{2}}-
10}$ ………………..(i)
Now, let’s try rationalising equation (i).
For that, let’s multiply the numerator and the denominator with the conjugate of whatever part of the
fraction has the radicals. Since we have radicals in the numerator, we’ll multiply the numerator and
denominator by its conjugate, which $=\sqrt{7+2x}+(\sqrt{5}+\sqrt{2})$
Doing so, we get :
$\Rightarrow \underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{\sqrt{7+2x}-\left( \sqrt{5}+\sqrt{2}
\right)}{{{x}^{2}}-10}\times \dfrac{\sqrt{7+2x}+\left( \sqrt{5}+\sqrt{2} \right)}{\sqrt{7+2x}+\left(
\sqrt{5}+\sqrt{2} \right)}$
Now, above equation numerator looks similar to the identity,
$\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$, where $a=\sqrt{7+2x},b=\sqrt{5}+\sqrt{2}$
So what we could do here is, apply the identity mentioned above. Doing so, we get :

\[\Rightarrow \underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{{{\left( \sqrt{7+2x} \right)}^{2}}-{{\left(
\sqrt{5}+\sqrt{2} \right)}^{2}}}{({{x}^{2}}-10)\left( \sqrt{7+2x}+\left( \sqrt{5}+\sqrt{2} \right) \right)}\]
Now, let’s try further simplifying the above equation. After solving the above equation further, we get;
$\to \underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{\left( 7+2x \right)-\left( 5+2+2\sqrt{10}
\right)}{\left( {{x}^{2}}-10 \right)\left( \sqrt{7+2x}+\sqrt{5}+\sqrt{2} \right)}$ (Using property:
\[\underset{{}}{\mathop{{{\left( a+b \right)}^{2}}}}\,={{a}^{2}}+{{b}^{2}}+2ab\] )

Now, on opening the brackets in the numerator and solving it, we get :
$\Rightarrow \underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{{7}+2x-{7}-
2\sqrt{10}}{\left( {{x}^{2}}-10 \right)\left( \sqrt{7+2x}+\sqrt{5}+\sqrt{2} \right)}$
Taking $2$ common from the terms in the numerator, we get :
$\Rightarrow \underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{2{\left( x-\sqrt{10}
\right)}}{{\left( x-\sqrt{10} \right)}\left( x+\sqrt{10} \right)\left( \sqrt{7+2x}+\sqrt{5}+\sqrt{2}
\right)}$……………….(ii)
Now, let’s put the \[\underset{x\to \sqrt{10}}{\mathop{\lim }}\,\] in the equation (ii). Doing so, we get :
$\Rightarrow \dfrac{2}{\left( \sqrt{10}+\sqrt{10} \right)\left( \sqrt{7+2\sqrt{10}}+\sqrt{5}+\sqrt{2}
\right)}$
$\Rightarrow \dfrac{{{2}}}{{2}\sqrt{10}\left( \left( \sqrt{7+2\sqrt{10}}
\right)+\sqrt{5}+\sqrt{2} \right)}$
$\Rightarrow \dfrac{1}{\sqrt{10}\left( \sqrt{7+2\sqrt{10}}+\sqrt{5}+\sqrt{2} \right)}$
Hence, the solution is $\dfrac{1}{\sqrt{10}\left( \sqrt{7+2\sqrt{10}}+\sqrt{5}+\sqrt{2}
\right)}$
Note: We, here have used rationalisation followed by substitution to solve the limit. We could also use
L’ Hopital Rule, because the limit becomes a $\dfrac{0}{0}$ form on putting $x=\sqrt{10}$ in the limit.
Note that,
$\sqrt{7+2\sqrt{10}}=\sqrt{{{(\sqrt{5})}^{2}}+{{(\sqrt{2})}^{2}}+2.\sqrt{2}.\sqrt{5}}=\sqrt{{{\left(
\sqrt{5}+\sqrt{2} \right)}^{2}}}=\sqrt{5}+\sqrt{2}$, and this is what makes the numerator zero on putting
$x=\sqrt{10}$. Thus, you could’ve used L’ Hopital Rule over here as well.