lim_(x rarr-oo)(6x^(2)-x)/(sqrt(9x^(4)+7x^(3)))=

Simplify expression by dividing: To find the limit of the given function as x approaches negative infinity, we can simplify the expression by dividing the numerator and the denominator by the highest power of x in the denominator, which is x^2.Remove terms as x approaches negative infinity: Divide both the numerator and the denominator by x^2: lim_(x -> -∞) [(6x^2 - x) / x^2] / [sqrt(9x^4 + 7x^3) / x^2] = lim_(x -> -∞) [6 - 1/x] / [sqrt(9 + 7/x)]Simplify expression further: As x approaches negative infinity, the terms 1/x and 7/x will approach 0. Therefore, we can simplify the expression further by removing these terms: lim_(x -> -∞) [6 - 0] / [sqrt(9 + 0)] = lim_(x -> -∞) 6 / sqrt(9)Simplify constant expression: Now we can simplify the constant expression: 6 / sqrt(9) = 6 / 3 = 2

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Math Problems

Grade 8

Relationship between squares and square roots

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\lim _{x \rightarrow-\infty} \frac{6 x^{2}-x}{\sqrt{9 x^{4}+7 x^{3}}}=

Simplify expression by dividing: To find the limit of the given function as $x$ approaches negative infinity, we can simplify the expression by dividing the numerator and the denominator by the highest power of $x$ in the denominator, which is $x^2$ .
Remove terms as $x$ approaches negative infinity: Divide both the numerator and the denominator by $x^2$ :
$\lim_{x \to -\infty} \left[\frac{6x^2 - x}{x^2}\right] / \left[\frac{\sqrt{9x^4 + 7x^3}}{x^2}\right]$
= $\lim_{x \to -\infty} \left[\frac{6 - \frac{1}{x}}{\sqrt{9 + \frac{7}{x}}}\right]$
Simplify expression further: As $x$ approaches negative infinity, the terms $\frac{1}{x}$ and $\frac{7}{x}$ will approach $0$ . Therefore, we can simplify the expression further by removing these terms: $\newline$ $\lim_{x \to -\infty} \frac{6 - 0}{\sqrt{9 + 0}}$ $\newline$ $= \lim_{x \to -\infty} \frac{6}{\sqrt{9}}$
Simplify constant expression: Now we can simplify the constant expression: $6 / \sqrt{9} = 6 / 3 = 2$