Find the limit as x approaches negative infinity. lim_(x rarr-oo)(3x)/(sqrt(16x^(2)-9x))=

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Find the limit as  x approaches negative infinity.  lim_(x rarr-oo)(3x)/(sqrt(16x^(2)-9x))=

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Identify highest power of x: Identify the highest power of x in the denominator. In the expression sqrt(16x^2-9x), the highest power of x is x^2 inside the square root. This means that the square root behaves like sqrt(x^2) for large values of x.Factor out highest power of x: Factor out the highest power of x from the square root in the denominator. We can write sqrt(16x^2-9x) as sqrt(x^2(16-9/x)). For large values of x, the term 9/x approaches 0, so we can approximate the expression as sqrt(16x^2).Simplify expression by dividing: Simplify the expression by dividing both the numerator and the denominator by x. We get (3x)/(sqrt(16x^2)) = (3x)/(4|x|). Since we are considering the limit as x approaches negative infinity, |x| is equal to -x. Therefore, the expression simplifies to (3x)/(4(-x)).Simplify expression further: Simplify the expression further. The x's cancel out, and we are left with -3/4, since there is a negative sign from the |x| when x is negative.Conclude the limit: Conclude the limit. The limit of the expression as x approaches negative infinity is -3/4.

Find the limit as $x$ approaches negative infinity. $\newline$ $\lim _{x \rightarrow-\infty} \frac{3 x}{\sqrt{16 x^{2}-9 x}}=$

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Find the limit as x x x approaches negative infinity.\newlinelim⁡x→−∞3x16x2−9x= \lim _{x \rightarrow-\infty} \frac{3 x}{\sqrt{16 x^{2}-9 x}}= x→−∞lim​16x2−9x​3x​=

Find the limit as $x$ approaches negative infinity. $\newline$ $\lim _{x \rightarrow-\infty} \frac{3 x}{\sqrt{16 x^{2}-9 x}}=$