Math Doubts

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

$x$ is a literal. $\sqrt{3x}-3$ and $\sqrt{2x-4}-\sqrt{2}$ are two algebraic expressions. The quotient of them is formed an algebraic function. The value of the function is required to find when the value of $x$ tends to $3$.

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


Evaluation of the function

Firstly, let us try the substitution method to study the functionality of the function as the limit $x$ approaches $3$.

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

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

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

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

The value is indeterminate. So, the limit problem should be solved in another method.


Try Rationalizing method

The numerator and denominator become zero as $x$ approaches $3$. So, rationalize the numerator by its rationalizing factor and also rationalise the denominator by its rationalising factor.

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

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

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

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

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

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

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

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

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

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

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

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

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


Evaluation of the function

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

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

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

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

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

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

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

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

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

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

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

It is the required solution for this limits problem of the calculus mathematics.

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