Question

1. \\( \lim _ { x \

ightarrow 3 } \frac { x ^ { 3 } - 5 x ^ { 2 } + 7 x - 3 } { x ^ { 2 } - 7 x + 12 } \\)

Explanation:

Step1: Factor numerator and denominator

For numerator \(x^{3}-5x^{2}+7x - 3\), by synthetic division or trial - and - error, we find \(x^{3}-5x^{2}+7x - 3=(x - 3)(x^{2}-2x + 1)=(x - 3)(x - 1)^{2}\).
For denominator \(x^{2}-7x + 12=(x-3)(x - 4)\).
So, \(\lim_{x
ightarrow3}\frac{x^{3}-5x^{2}+7x - 3}{x^{2}-7x + 12}=\lim_{x
ightarrow3}\frac{(x - 3)(x - 1)^{2}}{(x - 3)(x - 4)}\).

Step2: Simplify the function

Cancel out the common factor \((x - 3)\) (for \(x
eq3\)), we get \(\lim_{x
ightarrow3}\frac{(x - 1)^{2}}{x - 4}\).

Step3: Substitute \(x = 3\)

Substitute \(x=3\) into \(\frac{(x - 1)^{2}}{x - 4}\), we have \(\frac{(3 - 1)^{2}}{3 - 4}=\frac{4}{-1}=-4\).

Answer:

\(-4\)