$x$ is a literal and its represents the value of limit of a function. The literal formed two functions and and the faction of them is required to find mathematical if the value of $x$ tends to $4$.
$\large \displaystyle \lim_{x \,\to\, 4}$ $\dfrac{3-\sqrt{5+x}}{x-4}$
Firstly, test this algebraic function when $x$ tends to $4$.
$\implies \dfrac{3-\sqrt{5+4}}{x-4}$ $=$ $\dfrac{3-\sqrt{9}}{4-4}$
$= \dfrac{3-3}{4-4}$
$= \dfrac{0}{0}$
The algebraic function became indeterminate when the value of $x$ approaches to $0$. Hence, the limit problem should be simplified in alternative method.
The numerator is in radical form. So, multiply both numerator and denominator by its rationalizing factor.
$=$ $\large \displaystyle \lim_{x \,\to\, 4}$ $\dfrac{3-\sqrt{5+x}}{x-4}$ $\times$ $\dfrac{3+\sqrt{5+x}}{3+\sqrt{5+x}}$
$=$ $\large \displaystyle \lim_{x \,\to\, 4}$ $\dfrac{(3-\sqrt{5+x})(3+\sqrt{5+x})}{(x-4)(3+\sqrt{5+x})}$
$=$ $\large \displaystyle \lim_{x \,\to\, 4}$ $\dfrac{{(3)}^2-{(\sqrt{5+x})}^2}{(x-4)(3+\sqrt{5+x})}$
$=$ $\large \displaystyle \lim_{x \,\to\, 4}$ $\dfrac{9-(5+x)}{(x-4)(3+\sqrt{5+x})}$
$=$ $\large \displaystyle \lim_{x \,\to\, 4}$ $\dfrac{9-5-x}{(x-4)(3+\sqrt{5+x})}$
$=$ $\large \displaystyle \lim_{x \,\to\, 4}$ $\dfrac{4-x}{(x-4)(3+\sqrt{5+x})}$
$=$ $\large \displaystyle \lim_{x \,\to\, 4}$ $\dfrac{-(x-4)}{(x-4)(3+\sqrt{5+x})}$
$=$ $\large \displaystyle \lim_{x \,\to\, 4}$ $\require{cancel} \dfrac{-\cancel{(x-4)}}{\cancel{(x-4)}(3+\sqrt{5+x})}$
Substitute $x = 4$ to determine the value of the algebraic function when $x$ tends to $4$.
$=$ $\large \displaystyle \lim_{x \,\to\, 4}$ $\dfrac{-1}{3+\sqrt{5+x}}$
$=$ $\dfrac{-1}{3+\sqrt{5+4}}$
$=$ $\dfrac{-1}{3+\sqrt{9}}$
$=$ $\dfrac{-1}{3+3}$
$=$ $\dfrac{-1}{6}$
$\therefore \,\,\,\,\,\,$ $\large \displaystyle \lim_{x \,\to\, 4}$ $\dfrac{3-\sqrt{5+x}}{x-4}$ $=$ $-\dfrac{1}{6}$
Therefore, it is the required solution for this limit problem in mathematics.
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