Question

${\lim _{x \to 2}}\dfrac{{\sqrt {1\sqrt {2 + x} } - \sqrt 3 }}{{x - 2}}$ is equal to:
A. $\dfrac{1}{{8\sqrt 3 }}$
B. $\dfrac{1}{{\sqrt 3 }}$
C. $8\sqrt 3 $
D. $\sqrt 3 $

Answer 1

Hint: A real-valued function's limit with regard to the parameter $x$ can be defined as follows:$\mathop {\lim }\limits_{x \to p} f(x) = L$. Here, the term $\lim $ in the equation above refers to the limit and a right arrow is used to indicate that the real-valued function$f(x)$ tends to reach the limit $L$ as $x$ goes to $p$.
To solve the given equation, we need to rationalize the equation and then apply the limit.

Formula Used:
The rule of rationalization is given by: The denominator contains a radical form as in the fraction $\dfrac{a}{{\sqrt b }}$. Here, the radical form must be multiplied and divided by $\sqrt b $ and simplified further.

Complete step by step Solution:
In the question, the equation is given by:
${\lim _{x \to 2}}\dfrac{{\sqrt {1\sqrt {2 + x} } - \sqrt 3 }}{{(x - 2)}}$
First, Divide and multiply the equation by $\sqrt {1\sqrt {2 + x} } + \sqrt 3 $ to rationalize the given equation, we have:
${\lim _{x \to 2}}\dfrac{{\sqrt {1\sqrt {2 + x} } - \sqrt 3 }}{{x - 2}} \times \dfrac{{\sqrt {1\sqrt {2 + x} } + \sqrt 3 }}{{\sqrt {1\sqrt {2 + x} } + \sqrt 3 }} \\$
 $\Rightarrow {\lim _{x \to 2}}\dfrac{{\sqrt {1\sqrt {2 + x} } - \sqrt 3 \times \sqrt {1\sqrt {2 + x} } + \sqrt 3 }}{{(x - 2)(\sqrt {1\sqrt {2 + x} } + \sqrt 3 )}} \\$
Now, expand the above equation, then:
${\lim _{x \to 2}}\dfrac{{1 + \sqrt {2 + x} - 3}}{{(x - 2)(\sqrt {1\sqrt {2 + x} } + \sqrt 3 )}} \\$
${\lim _{x \to 2}}\dfrac{{\sqrt {2 + x} - 2}}{{(x - 2)(\sqrt {1\sqrt {2 + x} } + \sqrt 3 )}} \\$
Multiplying and dividing the obtained equation by $\sqrt {2 + x} + 2$ to rationalize the equation again:
${\lim _{x \to 2}}\dfrac{{\sqrt {2 + x} - 2}}{{(x - 2)(\sqrt {1\sqrt {2 + x} } + \sqrt 3 )}} \times \dfrac{{\sqrt {2 + x} + 2}}{{\sqrt {2 + x} + 2}} \\$
$\Rightarrow {\lim _{x \to 2}}\dfrac{{(\sqrt {2 + x} - 2)(\sqrt {2 + x} + 2)}}{{(x - 2)(\sqrt {1\sqrt {2 + x} } + \sqrt 3 )(\sqrt {2 + x} + 2)}} \\$
Similarly, expand the above equation, then:
${\lim _{x \to 2}}\dfrac{{{{(\sqrt {2 + x} )}^2} - {2^2}}}{{(x - 2)(\sqrt {1\sqrt {2 + x} } + \sqrt 3 )(\sqrt {2 + x} + 2)}} \\$
${\lim _{x \to 2}}\dfrac{{x - 2}}{{(x - 2)(\sqrt {1\sqrt {2 + x} } + \sqrt 3 )(\sqrt {2 + x} + 2)}} \\$
Now, canceling out $(x - 2)$from numerator and denominator, thus:
${\lim _{x \to 2}}\dfrac{{(x - 2)}}{{(x - 2)(\sqrt {1\sqrt {2 + x} } + \sqrt 3 )(\sqrt {2 + x} + 2)}} \\$
$\Rightarrow {\lim _{x \to 2}}\dfrac{1}{{(\sqrt {1\sqrt {2 + x} } + \sqrt 3 )(\sqrt {2 + x} + 2)}} \\$
Apply the limit which is $x$tends to $2$, then we obtain:
$\Rightarrow {\lim _{x \to 2}}\dfrac{1}{{(\sqrt {1 + \sqrt {2 + (2)} } + \sqrt 3 )(\sqrt {2 + (2)} + 2)}} \\$
$\Rightarrow \dfrac{1}{{(\sqrt {1 + \sqrt 4 } + \sqrt 3 )(\sqrt 4 + 2)}} \\$
 $\Rightarrow \dfrac{1}{{(2\sqrt 3 )(2 + 2)}} \\$
 $\Rightarrow \dfrac{1}{{8\sqrt 3 }} \\$

Hence, the correct option is (A).

Note: It should be noted that if the limit of a function is given in a quotient form then we can determine the expression by using the factoring method. Under this method, we need to reduce the numerator and denominator of the equation to their factors completely and then simplify it by dividing the numerator and the denominator by any factor which is common to both.