Answer:
See below.
Step-by-step explanation:
We have:
[tex]\displaystyle \lim_{x\to \-8}\left(\frac{1}{1+x}- \frac{1}{9}\right)[/tex]
Simply plug in each value:
When x = 7.9, [tex]\displaystyle \frac{1}{1+x}- \frac{1}{9} \approx 0.0012[/tex]
When x = 7.99, [tex]\displaystyle \frac{1}{1+x}- \frac{1}{9} \approx0.0001[/tex]
When x = 7.999, [tex]\displaystyle \frac{1}{1+x}- \frac{1}{9} \approx0.0000[/tex]
When x = 8.001, [tex]\displaystyle \frac{1}{1+x}- \frac{1}{9} \approx-0.0000[/tex]
When x = 8.01, [tex]\displaystyle \frac{1}{1+x}- \frac{1}{9} \approx-0.0001[/tex]
And when x = 8.1, [tex]\displaystyle \frac{1}{1+x}- \frac{1}{9} \approx-0.0012[/tex]
From this pattern, we can conclude that:
[tex]\displaystyle \lim_{x\to \-8}\left(\frac{1}{1+x}- \frac{1}{9}\right) = 0[/tex]
Since as the limit approaches 8, the value gets smaller and smaller and approaches zero.
Graphing this, we can confirm our answer.
