Question

use a graph or a table to find the limit.

$$\lim_{x \to 1^{+}} \frac{-3}{x^{2}-1}$$

$$\lim_{x \to 1^{+}} \frac{-3}{x^{2}-1}=\square$$ (simplify your answer.)

Explanation:

Step1: Analyze the denominator as \(x\to1^{+}\)

When \(x\to1^{+}\), \(x^{2}-1=(x - 1)(x + 1)\to0^{+}\) (since \(x-1\to0^{+}\) and \(x + 1\to2\)).

Step2: Analyze the limit of the fraction

We have \(\lim_{x\to1^{+}}\frac{-3}{x^{2}-1}\). The numerator is \(-3\) (a non - zero constant), and the denominator \(x^{2}-1\to0^{+}\) as \(x\to1^{+}\).
By the rule of limits \(\lim_{x\to a}\frac{c}{f(x)}\) where \(c
eq0\) and \(\lim_{x\to a}f(x)=0^{+}\), we know that \(\lim_{x\to1^{+}}\frac{-3}{x^{2}-1}=-\infty\)

Answer:

\(-\infty\)