We are asked to find the left and right hand limits of the given function.
1. Left-hand limit:
[tex]\lim_{n\to1^-}\;\frac{|7x-7|}{1-x}[/tex]
The numerator is negative when we substitute -1, so the absolute value will be evaluated as a negative expression.
[tex]\lim_{n\to1^-}\;\frac{-(7x-7)}{1-x}=\frac{-7x+7}{1-x}[/tex]
Now, factor the numerator
[tex]\lim_{n\to1^-}\;\frac{-7x+7}{1-x}=\frac{7(-x+1)}{1-x}=\frac{7(1-x)}{1-x}=7[/tex]
So, the left-hand limit is equal to 7
2. Right-hand limit:
[tex]\lim_{n\to1+}\;\frac{|7x-7|}{1-x}[/tex]
The numerator is positive when we substitute +1, so the absolute value will be evaluated as a positive expression.
[tex]\lim_{n\to1+}\;\frac{|7x-7|}{1-x}=\frac{7x-7}{1-x}[/tex]
Now, factor the numerator and the denominator.
[tex]\lim_{n\to1+}\;\frac{7x-7}{1-x}=\frac{7(x-1)}{1-x}=\frac{7(x-1)}{-(x-1)}=-7[/tex]
So, the right-hand limit is equal to -7
