Answer
$$\frac{1}{8}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \left( {4x - \sqrt {16{x^2} - x} } \right) \cr
& {\text{Rationalizing}} \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{4x - \sqrt {16{x^2} - x} }}{1} \times \frac{{4x + \sqrt {16{x^2} - x} }}{{4x + \sqrt {16{x^2} - x} }}} \right] \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{{\left( {4x} \right)}^2} - {{\left( {\sqrt {16{x^2} - x} } \right)}^2}}}{{4x + \sqrt {16{x^2} - x} }}} \right) \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{16{x^2} - 16{x^2} + x}}{{4x + \sqrt {16{x^2} - x} }}} \right) \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to \infty } \left( {\frac{x}{{4x + \sqrt {16{x^2} - x} }}} \right) \cr
& {\text{Divide the numerator and denominator by }}x \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\frac{x}{x}}}{{\frac{{4x}}{x} + \sqrt {\frac{{16{x^2}}}{{{x^2}}} - \frac{x}{{{x^2}}}} }}} \right) \cr
& {\text{ = }}\mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{{4 + \sqrt {16 - \frac{1}{x}} }}} \right) \cr
& {\text{Evaluate the limit}} \cr
& {\text{ = }}\frac{1}{{4 + \sqrt {16 - 0} }} \cr
& = \frac{1}{8} \cr} $$
