Find the limit as x approaches negative infinity. lim_(x rarr-oo)(sqrt(16x^(6)-x^(2)))/(6x^(3)+x^(2))=

Identify highest power of x: Identify the highest power of x in the numerator and the denominator. The highest power of x in the numerator is x^6 (from the term 16x^6), and the highest power of x in the denominator is x^3 (from the term 6x^3).Divide by highest power: Divide every term by x^3, the highest power of x in the denominator. We get (sqrt(16x^6/x^3 - x^2/x^3))/(6x^3/x^3 + x^2/x^3) = (sqrt(16x^3 - 1/x))/(6 + 1/x).Simplify expression inside square root: Simplify the expression inside the square root. Since x is approaching negative infinity, 1/x approaches 0. Therefore, the term -1/x inside the square root becomes negligible, and we can approximate the expression as (sqrt(16x^3))/(6 + 1/x).Simplify square root: Simplify the square root. The square root of 16x^3 is 4x^(3/2). So the expression simplifies to (4x^(3/2))/(6 + 1/x).Approach negative infinity: As x approaches negative infinity, 1/x approaches 0. The term 1/x in the denominator becomes negligible, so the expression simplifies to (4x^(3/2))/6.Further simplify expression: Simplify the expression further. Since x^(3/2) is the same as (x^3)^(1/2), we can rewrite the expression as 4/(6x^(3/2)).

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Find the limit as
x approaches negative infinity.

lim_(x rarr-oo)(sqrt(16x^(6)-x^(2)))/(6x^(3)+x^(2))=

Find the limit as $x$ approaches negative infinity. $\newline$ $\lim _{x \rightarrow-\infty} \frac{\sqrt{16 x^{6}-x^{2}}}{6 x^{3}+x^{2}}=$

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Algebra 1

Compare linear and exponential growth

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Q. Find the limit as

x

approaches negative infinity.

\newline

\lim _{x \rightarrow-\infty} \frac{\sqrt{16 x^{6}-x^{2}}}{6 x^{3}+x^{2}}=

Identify highest power of x: Identify the highest power of $x$ in the numerator and the denominator. $\newline$ The highest power of $x$ in the numerator is $x^6$ (from the term $16x^6$ ), and the highest power of $x$ in the denominator is $x^3$ (from the term $6x^3$ ).
Divide by highest power: Divide every term by $x^3$ , the highest power of $x$ in the denominator. $\newline$ We get $(\sqrt{\frac{16x^6}{x^3} - \frac{x^2}{x^3}})/(\frac{6x^3}{x^3} + \frac{x^2}{x^3}) = (\sqrt{16x^3 - \frac{1}{x}})/(6 + \frac{1}{x})$ .
Simplify expression inside square root: Simplify the expression inside the square root. $\newline$ Since $x$ is approaching negative infinity, $\frac{1}{x}$ approaches $0$ . Therefore, the term $-\frac{1}{x}$ inside the square root becomes negligible, and we can approximate the expression as $\frac{\sqrt{16x^3}}{6 + \frac{1}{x}}$ .
Simplify square root: Simplify the square root. $\newline$ The square root of $16x^3$ is $4x^{\frac{3}{2}}$ . So the expression simplifies to $\frac{4x^{\frac{3}{2}}}{6 + \frac{1}{x}}$ .
Approach negative infinity: As $x$ approaches negative infinity, $\frac{1}{x}$ approaches $0$ . The term $\frac{1}{x}$ in the denominator becomes negligible, so the expression simplifies to $\frac{4x^{(3/2)}}{6}$ .
Further simplify expression: Simplify the expression further. $\newline$ Since $x^{(3/2)}$ is the same as $(x^3)^{(1/2)}$ , we can rewrite the expression as $\frac{4}{6x^{(3/2)}}$ .