limx→3^(-)(x^3-3x^2-4x+12)/(2x^3-7x^2+2x+3)=

Correct answer: (C) 5/14. Given, L=limx→3^(-)(x^3-3x^2-4x+12)/(2x^3-7x^2+2x+3) Checking the form we get, L=0/0 form at x=3 So, applying L-hospital rule in limit we get, L=limx→3^(-)(3x^2-6x-4)/(6x^2-14x+2) L=(3×3^2-6×3-4)/(6×3^2-14×3+2)=5/14

Q105 / 160 · TS EAMCET 2022 (18 Jul Shift 1) · Mathematics

$\lim_{x \to 3^{-}} \frac{x^{3} - 3 x^{2} - 4 x + 12}{2 x^{3} - 7 x^{2} + 2 x + 3} =$

A
$0$
B
$\infty$
Correct answer
C
$\frac{5}{14}$
D
$\frac{6}{13}$

Solution

Given,
$L = \lim_{x \to 3^{-}} \frac{x^{3} - 3 x^{2} - 4 x + 12}{2 x^{3} - 7 x^{2} + 2 x + 3}$
Checking the form we get, $L = \frac{0}{0} form at x = 3$
So, applying L-hospital rule in limit we get,
$L = \lim_{x \to 3^{-}} \frac{3 x^{2} - 6 x - 4}{6 x^{2} - 14 x + 2}$
$L = \frac{3 \times 3^{2} - 6 \times 3 - 4}{6 \times 3^{2} - 14 \times 3 + 2} = \frac{5}{14}$

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

limx→3-x3-3x2-4x+122x3-7x2+2x+3=

Solution

Relevant Questions

$\lim_{x \to 3^{-}} \frac{x^{3} - 3 x^{2} - 4 x + 12}{2 x^{3} - 7 x^{2} + 2 x + 3} =$