# Lim X → 1 X 4 − 3 X 3 + 2 X 3 − 5 X 2 + 3 X + 1 - Mathematics

$\lim_{x \to 1} \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1}$

#### Solution

p(x) = x4$-$  3x3 + 2
p(1) = 1$-$3 + 2
= 0
Now,

$\left( x - 1 \right)$ is a factor of p(x).

q(x) = x$-$ 5x2 + 3x + 1
q(1) = 1$-$5 + 3 + 1
= 0

$\text{ Now }, \left( x - 1 \right)$is a factor of q(x).

$\Rightarrow \lim_{x \to 1} \left[ \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1} \right]$
$= \lim_{x \to 1} \left[ \frac{\left( x - 1 \right)\left( x^3 - 2 x^2 - 2x - 2 \right)}{\left( x - 1 \right)\left( x^2 - 4x - 1 \right)} \right]$
$= \frac{(1 )^3 - 2 \left( 1 \right)^2 - 2\left( 1 \right) - 2}{\left( 1 \right)^2 - 4 \times 1 - 1}$
$= \frac{1 - 2 - 2 - 2}{1 - 4 - 1}$
$= \frac{- 5}{- 4}$
$= \frac{5}{4}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.3 | Q 25 | Page 23