Question

the limit found using either method is $limlimits_{x\to10}\frac{x - 10}{x^{2}-100}=square$ (simplify your answer.)

Explanation:

Step1: Factor the denominator

Use the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\). Here \(a=x\) and \(b = 10\), so \(x^{2}-100=(x + 10)(x - 10)\).
The limit becomes \(\lim_{x
ightarrow10}\frac{x - 10}{(x + 10)(x - 10)}\)

Step2: Cancel out the common factor

Cancel out the common factor \((x - 10)\) (for \(x
eq10\), when taking the limit \(x
ightarrow10\) but \(x
eq10\)).
We get \(\lim_{x
ightarrow10}\frac{1}{x + 10}\)

Step3: Substitute \(x = 10\)

Substitute \(x=10\) into \(\frac{1}{x + 10}\).
\(\frac{1}{10+10}=\frac{1}{20}\)

Answer:

\(\frac{1}{20}\)