Question

1. $\lim _{x \to 1} \frac{x^{2}-9 x+8}{x^{2}+7 x-8}$

Explanation:

Step1: Factor numerator and denominator

Factor \(x^{2}-9x + 8=(x - 1)(x - 8)\) and \(x^{2}+7x - 8=(x - 1)(x+8)\)
So, \(\lim_{x
ightarrow1}\frac{x^{2}-9x + 8}{x^{2}+7x - 8}=\lim_{x
ightarrow1}\frac{(x - 1)(x - 8)}{(x - 1)(x + 8)}\)

Step2: Simplify the function

Cancel out the common factor \((x - 1)\) (for \(x
eq1\)), we get \(\lim_{x
ightarrow1}\frac{x - 8}{x + 8}\)

Step3: Substitute \(x = 1\)

Substitute \(x=1\) into \(\frac{x - 8}{x + 8}\), we have \(\frac{1-8}{1 + 8}=\frac{-7}{9}\)

Answer:

\(-\frac{7}{9}\)