Answer
$ - \frac{1}{{27}}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{t \to - 1} {\left( {\frac{{2{t^5} - {t^4}}}{{5{t^2} + 4}}} \right)^3} \cr
& {\text{Use The power Law }}\mathop {\lim }\limits_{x \to a} {\left[ {f\left( x \right)} \right]^n} = {\left[ {\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right]^n}{\text{, }}n \in {\text{Z + }} \cr
& \mathop {\lim }\limits_{t \to - 1} {\left( {\frac{{2{t^5} - {t^4}}}{{5{t^2} + 4}}} \right)^3} = {\left[ {\mathop {\lim }\limits_{t \to - 1} \left( {\frac{{2{t^5} - {t^4}}}{{5{t^2} + 4}}} \right)} \right]^3} \cr
& \cr
& {\text{Use the Quotient Law}} \cr
& {\left[ {\mathop {\lim }\limits_{t \to - 1} \left( {\frac{{2{t^5} - {t^4}}}{{5{t^2} + 4}}} \right)} \right]^3} = {\left[ {\frac{{\mathop {\lim }\limits_{t \to - 1} \left( {2{t^5} - {t^4}} \right)}}{{\mathop {\lim }\limits_{t \to - 1} \left( {5{t^2} + 4} \right)}}} \right]^3} \cr
& {\text{Use the Sum and Difference Laws }}\left( {{\text{see page 95}}} \right) \cr
& \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) \pm g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} f\left( x \right) \pm \mathop {\lim }\limits_{x \to a} g\left( x \right) \cr
& {\left[ {\frac{{\mathop {\lim }\limits_{t \to - 1} \left( {2{t^5} - {t^4}} \right)}}{{\mathop {\lim }\limits_{t \to - 1} \left( {5{t^2} + 4} \right)}}} \right]^3} = {\left[ {\frac{{\mathop {\lim }\limits_{t \to - 1} \left( {2{t^5}} \right) - \mathop {\lim }\limits_{t \to - 1} \left( {{t^4}} \right)}}{{\mathop {\lim }\limits_{t \to - 1} \left( {5{t^2}} \right) + \mathop {\lim }\limits_{t \to - 1} \left( 4 \right)}}} \right]^3} \cr
& \cr
& {\text{Use the Constant Multiple Law }} \cr
& \mathop {\lim }\limits_{x \to a} \left[ {cf\left( x \right)} \right] = c\mathop {\lim }\limits_{x \to a} f\left( x \right) \cr
& = {\left[ {\frac{{2\mathop {\lim }\limits_{t \to - 1} \left( {{t^5}} \right) - \mathop {\lim }\limits_{t \to - 1} \left( {{t^4}} \right)}}{{5\mathop {\lim }\limits_{t \to - 1} \left( {{t^2}} \right) + \mathop {\lim }\limits_{t \to - 1} \left( 4 \right)}}} \right]^3} \cr
& \cr
& {\text{Use The power Law }}\mathop {\lim }\limits_{x \to a} {x^n} = {a^n}{\text{, }}n{\text{ is a positive integer }} \cr
& {\text{and the laws }}\mathop {\lim }\limits_{x \to a} x = a,{\text{ }}\mathop {\lim }\limits_{x \to a} c = c,{\text{ then}} \cr
& = {\left[ {\frac{{2{{\left( { - 1} \right)}^5} - {{\left( 1 \right)}^4}}}{{5{{\left( { - 1} \right)}^2} + \left( 4 \right)}}} \right]^3} \cr
& {\text{Simplify}} \cr
& = {\left( {\frac{{ - 3}}{9}} \right)^3} \cr
& = - \frac{1}{{27}} \cr} $$
