Answer
455.1k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)