Chapter 12.1, Problem 7E

To determine

To estimate: The limit of given function by numerically method.

Answer to Problem 7E

Limit of the function ${\underrightarrow{\lim }}_{t\to -3}\frac{x+3}{x^2-9}$ is $-\frac{1}{6}$ .

Explanation of Solution

Given: ${\underrightarrow{\lim }}_{t\to -3}\frac{x+3}{x^2-9}$

Consider $f\left(x\right)={\underrightarrow{\lim }}_{t\to -3}\frac{x+3}{x^2-9}$ be a given function.

And it has to find value of given points for estimate the limit.

  

  $f\left(x\right)={\underrightarrow{\lim }}_{t\to -3}\frac{x+3}{x^2-9}$

  $f\left(-3.1\right)={\underrightarrow{\lim }}_{t\to -3}\frac{-3.1+3}{9.61-9}=-\frac{1}{6.1}$

  $f\left(x\right)={\underrightarrow{\lim }}_{t\to -3}\frac{x+3}{x^2-9}$

  $f\left(-3.01\right)={\underrightarrow{\lim }}_{t\to -3}\frac{-3.01+3}{9.0601-9}=-\frac{1}{6.01}$

  $f\left(x\right)={\underrightarrow{\lim }}_{t\to -3}\frac{x+3}{x^2-9}$

  $f\left(-3.001\right)={\underrightarrow{\lim }}_{t\to -3}\frac{-3.001+3}{9.006001-9}=-\frac{1}{6.001}$

  $f\left(x\right)={\underrightarrow{\lim }}_{t\to -3}\frac{x+3}{x^2-9}$

  $f\left(-2.999\right)={\underrightarrow{\lim }}_{t\to -3}\frac{-2.999+3}{8.994001-9}=-\frac{1}{5.999}$

  $f\left(x\right)={\underrightarrow{\lim }}_{t\to -3}\frac{x+3}{x^2-9}$

  $f\left(-2.99\right)={\underrightarrow{\lim }}_{t\to -3}\frac{-2.99+3}{8.9401-9}=-\frac{1}{5.99}$

  $f\left(x\right)={\underrightarrow{\lim }}_{t\to -3}\frac{x+3}{x^2-9}$

  $f\left(-2.9\right)={\underrightarrow{\lim }}_{t\to -3}\frac{-2.9+3}{8.41-9}=-\frac{1}{5.9}$

Hence, limit of the function ${\underrightarrow{\lim }}_{t\to -3}\frac{x+3}{x^2-9}$ is $-\frac{1}{6}$ .