Solve the equation : $\lim _ { x \rightarrow 8 } \dfrac { \sqrt { 1 + \sqrt { 1 + x } } - 2 } { x - 8 } =?$

To resolve the zero over zero indeterminate form, we can apply L'Hopital's rule, which states that the limit of a quotient is the quotient of the derivatives. The derivative of the numerator $\sqrt{1 + \sqrt{1 + x}} - 2$ is $(1 / (2\sqrt{1 + \sqrt{1 + x}})) \cdot (1 / (2\sqrt{1 + x}))$. The derivative of the denominator $x - 8$ is simply 1. Substituting $x = 8$ into the derivative quotient yields $(1 / (2\sqrt{1 + 3})) \cdot (1 / (2\sqrt{9}))$, which evaluates to (1 / 4) \cdot (1 / 6). The final result is 1 / 24.

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