determine the limit shown below in simplest form. lim x→1 x² - 25 / x² + 4x - 5

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-25=(x + 5)(x - 5)$ using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. The denominator $x^{2}+4x - 5=(x + 5)(x - 1)$ by factoring the quadratic $ax^{2}+bx + c$ where $a = 1$, $b = 4$, $c=-5$ and finding two numbers that multiply to $ac=-5$ and add up to $b = 4$ (the numbers are 5 and - 1). So the function becomes $\frac{(x + 5)(x - 5)}{(x + 5)(x - 1)}$.

Step2: Simplify the function

Cancel out the common factor $(x + 5)$ (for $x\neq - 5$). The simplified function is $\frac{x - 5}{x - 1}$.

Step3: Evaluate the limit

Substitute $x = 1$ into the simplified function $\lim_{x\rightarrow1}\frac{x - 5}{x - 1}=\frac{1-5}{1 - 1}=\frac{-4}{0}$. Since the denominator is 0 and the numerator is non - zero, the limit does not exist.

Answer:

The limit does not exist.