Question

Evaluate limit: $\mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right)$.

Answer 1

The given question requires us to evaluate a limit. There are various methods and steps to evaluate a limit. Some of the common steps while solving limits involve rationalization and applying some basic results on frequently used limits. L’Hospital’s rule involves solving limits of indeterminate form by differentiating both numerator and denominator with respect to the variable separately and then applying the required limit. We will solve the given limit using L’Hospital’s rule.

Complete step by step answer:
We have to evaluate limit $\mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right)$ using L’Hospital’s rule. So, if we put the limit x tending to square root of two into the expression $\left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right)$, we get an indeterminate form limit. Hence, L’Hospital’s rule can be applied here to find the value of the concerned limit.

So, Applying L’Hospital’s rule, we have to differentiate both numerator and denominator with respect to the variable x separately and then apply the limit.Hence,
$\mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right)$
Now, we know the power rule of differentiation as $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{\left( {n - 1} \right)}}$. So, we get,
$\mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right) = \mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\dfrac{2}{{2\sqrt {3 + 2x} }}}}{2}} \right)$
We know that the derivative of a constant is zero. Also, cancelling the common factors in numerator and denominator, we get,
$\mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right) = \mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{1}{{2\sqrt {3 + 2x} }}} \right)$

Now, substituting in the value of variable as per the limit.
$\mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right) = \left( {\dfrac{1}{{2\sqrt {3 + 2\sqrt 2 } }}} \right)$
Simplifying the expression, we get,
$\mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right) = \left( {\dfrac{1}{{2\sqrt {{{\left( {\sqrt 2 } \right)}^2} + {{\left( 1 \right)}^2} + 2\left( 1 \right)\left( {\sqrt 2 } \right)} }}} \right)$
Now, we know the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$.
$\mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right) = \left( {\dfrac{1}{{2\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} }}} \right)$
Taking the square root, we get,
$\mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right) = \left( {\dfrac{1}{{2\left( {\sqrt 2 + 1} \right)}}} \right)$

Now, rationalizing the denominator by multiplying and dividing by $\left( {\sqrt 2 - 1} \right)$, we get,
$\mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right) = \dfrac{1}{{2\left( {\sqrt 2 + 1} \right)}} \times \dfrac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right)}}$
Using the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, we get,
$\mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right) = \dfrac{{\left( {\sqrt 2 - 1} \right)}}{{2\left( {{{\left( {\sqrt 2 } \right)}^2} - {1^2}} \right)}}$
$\Rightarrow \mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right)= \dfrac{{\left( {\sqrt 2 - 1} \right)}}{{2\left( {2 - 1} \right)}}$
Simplifying the expression, we get,
$\therefore \mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right)= \dfrac{{\left( {\sqrt 2 - 1} \right)}}{2}$

So, the value of the limit $\mathop {\lim }\limits_{x \to \sqrt 2 } \left( {\dfrac{{\sqrt {3 + 2x} - \left( {\sqrt 2 + 1} \right)}}{{{x^2} - 2}}} \right)$ is $\dfrac{{\left( {\sqrt 2 - 1} \right)}}{2}$.

Note: The given question is of indeterminate form. So, we can solve it using the L'Hospital rule. We must remember some basic derivatives and limits to solve such complex problems using the L'Hospital rule. We should take care of calculations while attempting these types of questions. We should also know the power rule and chain rule of differentiation in order to solve the given problem.