Question

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

Answer 1

p(x) = x⁴\[-\]  3x³ + 2
p(1) = 1\[-\]3 + 2
       = 0
^Now,

^{\[\left( x - 1 \right)\] is a factor of p(x).}

q(x) = x³ \[-\] 5x² + 3x + 1
q(1) = 1\[-\]5 + 3 + 1
       = 0

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

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