Answer
$$5$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{5{x^2} + 2x - 5}}{{\sqrt {{x^4} - 1} }} \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{5{x^2} + 2x - 5}}{{\sqrt {{x^4} - 1} }} = \frac{{5{{\left( \infty \right)}^2} + 2\left( \infty \right) - 5}}{{\sqrt {{{\left( \infty \right)}^4} - 1} }} = \frac{\infty }{\infty } \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{5{x^2} + 2x - 5}}{{\sqrt {{x^4} - 1} }} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{5{x^2}}}{{{x^2}}} + \frac{{2x}}{{{x^2}}} - \frac{5}{{{x^2}}}}}{{\sqrt {\frac{{{x^4}}}{{{x^4}}} - \frac{1}{{{x^4}}}} }} \cr
& {\text{ }} = \mathop {\lim }\limits_{x \to \infty } \frac{{5 + \frac{{2x}}{{{x^2}}} - \frac{5}{{{x^2}}}}}{{\sqrt {1 - \frac{1}{{{x^4}}}} }} \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{5 + \frac{2}{x} - \frac{5}{{{x^2}}}}}{{\sqrt {1 - \frac{1}{{{x^4}}}} }} = \frac{{5 + \frac{2}{\infty } - \frac{5}{{{{\left( \infty \right)}^2}}}}}{{\sqrt {1 - \frac{1}{{{{\left( \infty \right)}^4}}}} }} \cr
& {\text{ }} = \frac{5}{{\sqrt 1 }} = 5 \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{5{x^2} + 2x - 5}}{{\sqrt {{x^4} - 1} }} = 5 \cr} $$
