Question

Question: 27. Find the limit or show that it does not exist.

27. \(\mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^5} - x}}{{{x^4} + 3}}\)

Short answer

The value is \(\mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^5} - x}}{{{x^4} + 3}} = - \infty \).

Step-by-step solution

Simply the given limit 

Dive the rational function in the given limit,\(\mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^5} - x}}{{{x^4} + 3}}\), by\({x^4}\)and simplify as shown below:

\(\mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^5} - x}}{{{x^4} + 3}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( {2{x^5} - x} \right)/{x^4}}}{{\left( {{x^4} + 3} \right)/{x^4}}}\)

Simply further

Perform the division in numerator and denominator and apply necessary limit laws. So, simplify further as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^5} - x}}{{{x^4} + 3}} &= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( {2{x^5} - x} \right)/{x^4}}}{{\left( {{x^4} + 3} \right)/{x^4}}}\\ &= \mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^5}/{x^4} - x/{x^4}}}{{{x^4}/{x^4} + 3/{x^4}}}\\ &= \mathop {\lim }\limits_{x \to - \infty } \frac{{2x - 1/{x^3}}}{{1 + 3/{x^4}}}\end{aligned}\).

Finally, apply the limit

As \(x \to - \infty \), \(2x - 1/{x^3}\) approaches \( - \infty \) and \(1 + 3/{x^4}\) approaches 1. So, on applying the limit to the above equation, and simplifying we get:

\(\begin{aligned}\mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^5} - x}}{{{x^4} + 3}} &= \mathop {\lim }\limits_{x \to - \infty } \frac{{2x - 1/{x^3}}}{{1 + 3/{x^4}}}\\ &= \frac{{ - \infty }}{1}\\ &= - \infty \end{aligned}\)

Thus, \(\mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^5} - x}}{{{x^4} + 3}} = - \infty \).