Question

Find the limit or show that it does not exist.

25. \(\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1 + 4{x^6}} }}{{2 - {x^3}}}\)

Short answer

The value is \(\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1 + 4{x^6}} }}{{2 - {x^3}}} = - 2\).

Step-by-step solution

Simply the given limit

Dive the rational function in the given limit,\(\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1 + 4{x^6}} }}{{2{x^3}}}\), by\({x^3}\)and simplify as shown below:

\(\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1 + 4{x^6}} }}{{2 - {x^3}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1 + 4{x^6}} /{x^3}}}{{\left( {2 - {x^3}} \right)/{x^3}}}\)

Simply further

Perform the division in numerator and denominator and apply necessary limit laws.

As\({x^3} = \sqrt {{x^6}} \), for\(x > 0\), division can be performed in the square root, positively. So, simplify further as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1 + 4{x^6}} }}{{2 - {x^3}}} &= \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {\left( {1 + 4{x^6}} \right)/{x^6}} }}{{\left( {2 - {x^3}} \right)/{x^3}}}\\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1/{x^6} + 4} }}{{2/{x^3} - 1}}\\ &= \frac{{\mathop {\lim }\limits_{x \to \infty } \sqrt {1/{x^6} + 4} }}{{\mathop {\lim }\limits_{x \to \infty } \left( {2/{x^3} - 1} \right)}}\\ &= \frac{{\sqrt {\mathop {\lim }\limits_{x \to \infty } \left( {1/{x^6}} \right) + \mathop {\lim }\limits_{x \to \infty } 4} }}{{\mathop {\lim }\limits_{x \to \infty } \left( {2/{x^3}} \right) - \mathop {\lim }\limits_{x \to \infty } 1}}\end{aligned}\).

Finally, apply the limit

On applying the limit to the above equation, and simplifying we get:

\(\begin{aligned}\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1 + 4{x^6}} }}{{2 - {x^3}}} &= \frac{{\sqrt {\mathop {\lim }\limits_{x \to \infty } \left( {1/{x^6}} \right) + \mathop {\lim }\limits_{x \to \infty } 4} }}{{\mathop {\lim }\limits_{x \to \infty } \left( {2/{x^3}} \right) - \mathop {\lim }\limits_{x \to \infty } 1}}\\ &= \frac{{\sqrt {0 + 4} }}{{ - 1}}\\ &= \frac{2}{{ - 1}}\\ &= - 2\end{aligned}\)

Thus, \(\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1 + 4{x^6}} }}{{2 - {x^3}}} = - 2\).