Answer:
Explanations:
Given the limit of the function expressed as:
[tex]\begin{gathered} \lim _{n\to0}\frac{\sin8x}{x} \\ f(x)=\frac{\sin 8x}{x} \end{gathered}[/tex]
First, we need to create a table for the given values in the table:
If x = -0.1
[tex]\begin{gathered} f(-0.1)=\frac{\sin8(-0.1)}{-0.1} \\ f(-0.1)=\frac{\sin(-0.8)}{-0.1} \\ f(-0.1)=0.1396 \end{gathered}[/tex]
If x = -0.01
[tex]\begin{gathered} f(-0.01)=\frac{\sin8(-0.01)}{-0.01} \\ f(-0.01)=\frac{\sin(-0.08)}{-0.01} \\ f(-0.01)=0.1396 \end{gathered}[/tex]
If x = -0.001
[tex]\begin{gathered} f(-0.001)=\frac{\sin8(-0.001)}{-0.001} \\ f(-0.001)=\frac{\sin(-0.008)}{-0.008} \\ f(-0.001)=0.1396 \end{gathered}[/tex]
From the values above, we can conclude that the values of f(x) will all tend to be 0.1396 for the positives values of x
Therefore, we can conclude that as you approach the value 0 from the positive and negative directions, they approach the same value, hence the limit does exist.
