In this limit problem, a function is given and the function is formed by the involvement of two irrational functions $\sqrt{3x}-3$ and $\sqrt{2x-4}-\sqrt{2}$ in ratio form. The limit of the given irrational function has to evaluate as the value of $x$ approaches to $3$.

$\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}}$

The limit of the given irrational function can be calculated in two different methods. Let us learn each method in step by step for evaluating the limit of the function as $x$ tends to $3$.

Let us learn how to find the limit of the quotient of square root of $3x$ minus $3$ by the square root of $2x$ minus $4$ minus square root of $2$ fundamentally as $x$ approaches $3$.

Let us try to evaluate the limit of this irrational function as $x$ approaches to $3$ by the direct substitution method.

$=\,\,\,$ $\dfrac{\sqrt{3(3)}-3}{\sqrt{2(3)-4}-\sqrt{2}}$

$=\,\,\,$ $\dfrac{\sqrt{9}-3}{\sqrt{6-4}-\sqrt{2}}$

$=\,\,\,$ $\dfrac{3-3}{\sqrt{2}-\sqrt{2}}$

$=\,\,\,$ $\dfrac{0}{0}$

It is evaluated that the limit of the given irrational function is indeterminate. So, the direct substitution method is not recommendable for finding the limit of the given irrational function. Hence, we have to think for an alternative method.

The value of the expression in the numerator is equal to zero as the value of $x$ is closer to $3$. In order to overcome this issue, multiply the numerator by its rationalizing factor.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}$ $\times$ $1$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}$ $\times$ $\dfrac{\sqrt{3x}+3}{\sqrt{3x}+3}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{(\sqrt{3x}-3)(\sqrt{3x}+3)}{(\sqrt{2x-4}-\sqrt{2})(\sqrt{3x}+3)}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{ {(\sqrt{3x})}^2-3^2}{(\sqrt{2x-4}-\sqrt{2})(\sqrt{3x}+3)}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{3x-9}{(\sqrt{2x-4}-\sqrt{2})(\sqrt{3x}+3)}$

The value of the expression in the denominator is also equal to zero as the value of $x$ tends to $3$. In order to solver this issue, multiply the denominator by its rationalizing factor.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{3x-9}{(\sqrt{2x-4}-\sqrt{2})(\sqrt{3x}+3)}$ $\times$ $1$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{3x-9}{(\sqrt{2x-4}-\sqrt{2})(\sqrt{3x}+3)}$ $\times$ $\dfrac{\sqrt{2x-4}+\sqrt{2}}{\sqrt{2x-4}+\sqrt{2}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{(3x-9)(\sqrt{2x-4}+\sqrt{2})}{(\sqrt{3x}+3)(\sqrt{2x-4}-\sqrt{2})(\sqrt{2x-4}+\sqrt{2})}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{(3x-9)(\sqrt{2x-4}+\sqrt{2})}{(\sqrt{3x}+3)({(\sqrt{2x-4})}^2-{(\sqrt{2})}^2)}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{(3x-9)(\sqrt{2x-4}+\sqrt{2})}{(\sqrt{3x}+3)(2x-4-2)}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{(3x-9)(\sqrt{2x-4}+\sqrt{2})}{(\sqrt{3x}+3)(2x-6)}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{(3x-9)(\sqrt{2x-4}+\sqrt{2})}{(2x-6)(\sqrt{3x}+3)}$

The second factor in both numerator and denominator are sum basis binomials in irrational form. Their values are not equal to zero as the value of $x$ approaches to $3$. Now, focus on simplifying the remaining factors.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{(3 \times x-3 \times 3)(\sqrt{2x-4}+\sqrt{2})}{(2 \times x-2 \times 3)(\sqrt{3x}+3)}$

Now, separate the common factor from terms in first factor of both numerator and denominator.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{3(x-3)(\sqrt{2x-4}+\sqrt{2})}{2(x-3)(\sqrt{3x}+3)}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\require{cancel} \dfrac{3\cancel{(x-3)}(\sqrt{2x-4}+\sqrt{2})}{2\cancel{(x-3)}(\sqrt{3x}+3)}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}$ $\dfrac{3(\sqrt{2x-4}+\sqrt{2})}{2(\sqrt{3x}+3)}$

The simplification process is completed and it is the time to evaluate the limit of function when $x$ approaches to $3$.

$=\,\,\,$ $\dfrac{3(\sqrt{2(3)-4}+\sqrt{2})}{2(\sqrt{3(3)}+3)}$

$=\,\,\,$ $\dfrac{3(\sqrt{6-4}+\sqrt{2})}{2(\sqrt{9}+3)}$

$=\,\,\,$ $\dfrac{3(\sqrt{2}+\sqrt{2})}{2(3+3)}$

$=\,\,\,$ $\dfrac{3(2\sqrt{2})}{2(6)}$

$=\,\,\,$ $\dfrac{6\sqrt{2}}{2 \times 6}$

$=\,\,\,$ $\require{cancel} \dfrac{\cancel{6}\sqrt{2}}{2 \times \cancel{6}}$

$=\,\,\,$ $\dfrac{\sqrt{2}}{2}$

$=\,\,\,$ $\dfrac{\sqrt{2}}{{(\sqrt{2})}^2}$

$=\,\,\,$ $\require{cancel} \dfrac{\cancel{\sqrt{2}}}{\cancel{{(\sqrt{2})}^2}}$

$=\,\,\,$ $\dfrac{1}{\sqrt{2}}$

$\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}}$

The expressions in the both numerator and denominator are defined in terms of $x$. Hence, differentiate each expression with respect to $x$ to implement the L’Hospital’s rule.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{d}{dx}(\sqrt{3x}-3)}{\dfrac{d}{dx}(\sqrt{2x-4}-\sqrt{2})}}$

Each algebraic expression in both numerator and denominator is formed by the difference of two expressions. The derivative of difference of the terms can be evaluated by the difference of their derivatives as per the difference rule of the derivatives.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{d}{dx}(\sqrt{3x})-\dfrac{d}{dx}(3)}{\dfrac{d}{dx}(\sqrt{2x-4})-\dfrac{d}{dx}(\sqrt{2})}}$

According to the derivative rule of a constant, the differentiation of a constant is zero.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{1}{2\sqrt{3x}} \times \dfrac{d}{dx}(3x)-0}{\dfrac{1}{2\sqrt{2x-4}} \times \dfrac{d}{dx}{(2x-4)}-0}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{1}{2\sqrt{3x}} \times \dfrac{d}{dx}(3x)}{\dfrac{1}{2\sqrt{2x-4}} \times \dfrac{d}{dx}{(2x-4)}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{1}{2\sqrt{3x}} \times \dfrac{d}{dx}(3x)}{\dfrac{1}{2\sqrt{2x-4}} \times \bigg(\dfrac{d}{dx}{(2x)}-\dfrac{d}{dx}{(4)}\bigg)}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{1}{2\sqrt{3x}} \times \dfrac{d}{dx}(3 \times x)}{\dfrac{1}{2\sqrt{2x-4}} \times \bigg(\dfrac{d}{dx}{(2 \times x)}-\dfrac{d}{dx}{(4)}\bigg)}}$

Now, separate the constant factors from the terms as per the constant multiple rule of the derivatives.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{1}{2\sqrt{3x}} \times 3 \times \dfrac{d}{dx}{(x)}}{\dfrac{1}{2\sqrt{2x-4}} \times \bigg(2 \times \dfrac{d}{dx}{(x)}-\dfrac{d}{dx}{(4)}\bigg)}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{1}{2\sqrt{3x}} \times 3 \times \dfrac{dx}{dx}}{\dfrac{1}{2\sqrt{2x-4}} \times \bigg(2 \times \dfrac{dx}{dx}-\dfrac{d}{dx}{(4)}\bigg)}}$

The derivative of the variable with respect to same variable is one as per the derivative rule of the variable with respect to same variable.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{1}{2\sqrt{3x}} \times 3 \times 1}{\dfrac{1}{2\sqrt{2x-4}} \times (2 \times 1-0)}}$

The L’Hopital’s rule is successfully used once. Now, let us simplify the irrational function.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{1}{2\sqrt{3x}} \times 3}{\dfrac{1}{2\sqrt{2x-4}} \times (2-0)}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{1}{2\sqrt{3x}} \times 3}{\dfrac{1}{2\sqrt{2x-4}} \times (2)}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{1}{2\sqrt{3x}} \times 3}{\dfrac{1}{2\sqrt{2x-4}} \times 2}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{1 \times 3}{2\sqrt{3x}}}{\dfrac{1 \times 2}{2\sqrt{2x-4}}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{3}{2\sqrt{3x}}}{\dfrac{1 \times \cancel{2}}{\cancel{2}\sqrt{2x-4}}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{\dfrac{3}{2\sqrt{3x}}}{\dfrac{1}{\sqrt{2x-4}}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \Bigg(\dfrac{3}{2\sqrt{3x}} \times \dfrac{\sqrt{2x-4}}{1}\Bigg)}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{3 \times \sqrt{2x-4}}{2\sqrt{3x} \times 1}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 3}{\normalsize \dfrac{3\sqrt{2x-4}}{2\sqrt{3x}}}$

The given irrational function is successfully simplified and now, let’s evaluate the limit of the simplified irrational function as $x$ approaches $3$ by the direct substitution.

$=\,\,\,$ $\dfrac{3\sqrt{2(3)-4}}{2\sqrt{3(3)}}$

$=\,\,\,$ $\dfrac{3\sqrt{6-4}}{2\sqrt{9}}$

$=\,\,\,$ $\dfrac{3\sqrt{2}}{2 \times 3}$

$=\,\,\,$ $\dfrac{3 \times \sqrt{2}}{2 \times 3}$

$=\,\,\,$ $\dfrac{\cancel{3} \times \sqrt{2}}{2 \times \cancel{3}}$

$=\,\,\,$ $\dfrac{\sqrt{2}}{2}$

$=\,\,\,$ $\dfrac{\sqrt{2}}{(\sqrt{2})^2}$

$=\,\,\,$ $\dfrac{1 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$

$=\,\,\,$ $\dfrac{1 \times \cancel{\sqrt{2}}}{\sqrt{2} \times \cancel{\sqrt{2}}}$

$=\,\,\,$ $\dfrac{1}{\sqrt{2}}$

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