A mathematical expression is formed in terms of literals $x$, $a$ and $b$. In this problem, $x$ is a variable but $a$ and $b$ are constants, and we have to evaluate the limit of the function as $x$ approaches $a$.

$\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{\sqrt{x-b}-\sqrt{a-b}}{x^2-a^2}}$

Now, let us try the basic mathematical approach the direct substitution method for evaluating the limit of the function as $x$ approaches $a$.

$=\,\,\,$ $\dfrac{\sqrt{a-b}-\sqrt{a-b}}{a^2-a^2}$

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

According to the direct substitution method, the limit of the given function is indeterminate as $x$ closer to $a$. Hence, we have to evaluate the limit of the function in another mathematical approach.

The expression in the numerator is an irrational function and it is also a main reason for the indeterminate form in this limit problem. So, the expression in the numerator should be rationalized.

So, let us try the rationalizing method for avoiding the indeterminate form.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \Bigg(\dfrac{\sqrt{x-b}-\sqrt{a-b}}{x^2-a^2}}$ $\times$ $\dfrac{\sqrt{x-b}+\sqrt{a-b}}{\sqrt{x-b}+\sqrt{a-b}}\Bigg)$

Now, focus on simplifying the mathematical expression for evaluating the limit of the given function. The product of the fractional functions can be simplified by the multiplication of fractions.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{\Big(\sqrt{x-b}-\sqrt{a-b}\Big)\Big(\sqrt{x-b}+\sqrt{a-b}\Big)}{\Big(x^2-a^2\Big)\Big(\sqrt{x-b}+\sqrt{a-b}\Big)}}$

The product of the expressions in the numerator can be simplified by the difference rule of squares.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{\Big(\sqrt{x-b}\Big)^2-\Big(\sqrt{a-b}\Big)^2}{\Big(x^2-a^2\Big)\Big(\sqrt{x-b}+\sqrt{a-b}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x-b-(a-b)}{\Big(x^2-a^2\Big)\Big(\sqrt{x-b}+\sqrt{a-b}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x-b-a+b}{\Big(x^2-a^2\Big)\Big(\sqrt{x-b}+\sqrt{a-b}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x-b+b-a}{\Big(x^2-a^2\Big)\Big(\sqrt{x-b}+\sqrt{a-b}\Big)}}$

$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x-\cancel{b}+\cancel{b}-a}{\Big(x^2-a^2\Big)\Big(\sqrt{x-b}+\sqrt{a-b}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x-a}{\Big(x^2-a^2\Big)\Big(\sqrt{x-b}+\sqrt{a-b}\Big)}}$

Now, factorize the difference of squares by the same difference of squares formula.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x-a}{\Big(x-a\Big)\Big(x+a\Big)\Big(\sqrt{x-b}+\sqrt{a-b}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{\cancel{x-a}}{\Big(\cancel{x-a}\Big)\Big(x+a\Big)\Big(\sqrt{x-b}+\sqrt{a-b}\Big)}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{1}{\Big(x+a\Big)\Big(\sqrt{x-b}+\sqrt{a-b}\Big)}}$

In the previous step, we have successfully simplified the mathematical expression and it cannot be simplified further. Therefore, we can now evaluate the limit of the function as $x$ tends to $a$. Now, the limit of the function can be evaluated by the direct substitution method.

$=\,\,\,$ $\dfrac{1}{\Big(a+a\Big)\Big(\sqrt{a-b}+\sqrt{a-b}\Big)}$

$=\,\,\,$ $\dfrac{1}{\Big(2a\Big)\Big(2\sqrt{a-b}\Big)}$

$=\,\,\,$ $\dfrac{1}{2a \times 2\sqrt{a-b}}$

$=\,\,\,$ $\dfrac{1}{4a\sqrt{a-b}}$

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