Answer

Verified

424.2k+ views

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.

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.

Recently Updated Pages

Assertion The resistivity of a semiconductor increases class 13 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE

What are the possible quantum number for the last outermost class 11 chemistry CBSE

Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

a Tabulate the differences in the characteristics of class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE

What organs are located on the left side of your body class 11 biology CBSE

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers