A calculus problem by Abdulrahman El Shafei

Calculus Level 3

lim ⁡ x → − ∞ 9 x 2 − 5 4 x + 3 = ? \large \lim_{ x\to -\infty }\dfrac { \sqrt { { 9x }^{ 2 }-5 } }{ 4x+3 } = \, ?


The answer is -0.75.

Abdulrahman El Shafei by Abdulrahman El Shafei , 23, Egypt

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2 solutions

L = lim ⁡ x → − ∞ 9 x 2 − 5 4 x + 3 Let u = − x = lim ⁡ u → ∞ 9 u 2 − 5 − 4 u + 3 Divide up and down by u = lim ⁡ u → ∞ 9 − 5 u 2 − 4 + 3 u = 9 − 4 = − 3 4 = − 0.75 \begin{aligned} L & = \lim_{x \to -\infty} \frac {\sqrt{9x^2-5}}{4x+3} & \small \color{#3D99F6} \text{Let }u = -x \\ & = \lim_{u \to \infty} \frac {\sqrt{9u^2-5}}{-4u+3} & \small \color{#3D99F6} \text{Divide up and down by }u \\ & = \lim_{u \to \infty} \frac {\sqrt{9-\frac 5{u^2}}}{-4+\frac 3u} \\ & = \frac {\sqrt{9}}{-4} \\ & = - \frac 34 = \boxed{-0.75} \end{aligned}

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Let x = − 99999 x=-99999 , then we have

9 ( − 99999 ) 2 − 5 4 ( − 99999 ) + 3 = − 0.75 \dfrac{\sqrt{9(-99999)^2-5}}{4(-99999)+3}=-0.75

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