limx→-∞(3|x|^3-x^2+2|x|-5)/(-5|x|^3+3x^2-2|x|+7)=

Correct answer: (D) (-3)/5. limx→-∞(3x^3-x^2+2x-5)/(-5x^3+3x^2-2x+7) Put x=-t =limt→∞(-3t^3-t^2-2t-5)/(5t^3+3t^2+2t+7) =limt→∞(-3-1t-2t^2-5t^3)/(5+3t+2t^2+7t^3) =(-3)/5, as limt→∞t^(-n)=0, where n∈N

Q105 / 160 · TS EAMCET 2021 (05 Aug Shift 1) · Mathematics

$\lim_{x \to - \infty} \frac{3 | x |^{3} - x^{2} + 2 | x | - 5}{- 5 | x |^{3} + 3 x^{2} - 2 | x | + 7} =$

A
$\frac{3}{5}$
B
$\frac{- 5}{7}$
C
$\frac{5}{7}$
Correct answer
D
$\frac{- 3}{5}$

Solution

$\lim_{x \to - \infty} \frac{3 {|x|}^{3} - x^{2} + 2 |x| - 5}{- 5 {|x|}^{3} + 3 x^{2} - 2 |x| + 7}$

Put $x = - t$

$= \lim_{t \to \infty} \frac{- 3 t^{3} - t^{2} - 2 t - 5}{5 t^{3} + 3 t^{2} + 2 t + 7}$

$= \lim_{t \to \infty} \frac{- 3 - \frac{1}{t} - \frac{2}{t^{2}} - \frac{5}{t^{3}}}{5 + \frac{3}{t} + \frac{2}{t^{2}} + \frac{7}{t^{3}}}$

$= \frac{- 3}{5}$ , as $\lim_{t \to \infty} t^{- n} = 0$ , where $n \in N$

Hence, the correct answer is option: 'D'.

limx→-∞3|x|3-x2+2|x|-5-5|x|3+3x2-2|x|+7=

Solution

Relevant Questions

$\lim_{x \to - \infty} \frac{3 | x |^{3} - x^{2} + 2 | x | - 5}{- 5 | x |^{3} + 3 x^{2} - 2 | x | + 7} =$