Question

question
determine the limit shown below in simplest form.

$$ lim _ { x ightarrow 1 } \frac { x ^ { 2 } - 25 } { x ^ { 2 } + 4 x - 5 } $$

Explanation:

Step1: Factor numerator and denominator

• Numerator: \(x^{2}-25=(x + 5)(x - 5)\) (using \(a^{2}-b^{2}=(a + b)(a - b)\) with \(a=x\) and \(b = 5\)).
• Denominator: \(x^{2}+4x - 5=(x + 5)(x - 1)\) (using \(x^{2}+(a + b)x+ab=(x + a)(x + b)\), here \(a = 5\), \(b=-1\) since \(5\times(-1)=-5\) and \(5+( - 1)=4\)).

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 \(x
eq - 5\), which is not relevant as \(x\to1\)). The simplified function is \(\frac{x - 5}{x - 1}\).

Step3: Substitute \(x = 1\) into the simplified function

\(\lim_{x
ightarrow1}\frac{x - 5}{x - 1}=\frac{1-5}{1 - 1}=\frac{-4}{0}\). Since the numerator is non - zero (\(-4
eq0\)) and the denominator approaches \(0\) as \(x\to1\), the limit does not exist.

Answer:

The limit \(\lim_{x
ightarrow1}\frac{x^{2}-25}{x^{2}+4x - 5}\) does not exist.