Question

use a graph or a table to find the limit. lim_{x→4^{+}}\frac{3}{x^{2}-16} lim_{x→4^{+}}\frac{3}{x^{2}-16}= (simplify your answer.)

Explanation:

Step1: Analyze the denominator

As \(x
ightarrow4^{+}\), consider \(y = x^{2}-16=(x - 4)(x + 4)\). When \(x
ightarrow4^{+}\), \(x-4>0\) and \(x - 4
ightarrow0\), and \(x + 4
ightarrow8\).

Step2: Analyze the whole - function limit

We have \(\lim_{x
ightarrow4^{+}}\frac{3}{x^{2}-16}=\lim_{x
ightarrow4^{+}}\frac{3}{(x - 4)(x + 4)}\). Since the numerator is a non - zero constant \(3\), and the denominator \((x - 4)(x + 4)\) approaches \(0\) from the positive side (because \(x
ightarrow4^{+}\) makes \(x - 4>0\) and \(x+4
ightarrow8\)), the value of the fraction \(\frac{3}{(x - 4)(x + 4)}\) approaches \(+\infty\).

Answer:

\(+\infty\)