Finding left-hand and right-hand limits of functions at given point?

[AA23]

Hello, I'm having hard time trying to go through this exercise. Please explain a) and b) so I can do the rest.
All help is appreciated!

 

[AA23]

Can someone please help with these? I can't seem to grasp how to do these no matter how many tutorials I watch.

 

[HallsofIvy]

First, are you clear what "one-sided limit" means? The two side limits, as x goes to a, are "x is to the right of a" and "x is to the left of a" or "x is larger than a" or "x is less than a".

Here you are taking the one sided limits as x goes to 1. "x to the right of 1" means x> 1 so x- 1> 0 and |x- 1|= x- 1. So, if x> 1, $\displaystyle \frac{|x- 1|}{x- 1}= \frac{x- 1}{x- 1}= 1$. What is the limit of that as x goes to 1 from the right?

"x to the left of 1" means x< 1 so x- 1< 0 and |x- 1|= -(x- 1). If x< 1, $\displaystyle \frac{|x- 1|}{x- 1}= -\frac{x- 1}{x- 1}= -1$. What is the limit of that as x goes to 1 from the left?


For the second, $\displaystyle \lim_{x\to 0} \frac{1}{x^{4/3}}- \frac{1}{(x- 2)^{1/3}}$
There no problem with the second fraction. At x= 0 that second fraction has value $\displaystyle \frac{1}{2^{1/3}}$. What happens to the first fraction? Obviously, as x goes to 0, we have a fraction with numerator 1 and denominator getting smaller and smaller. That means the fraction itself is going to "infinity" but it will be either positive or negative infinity depending upon whether the fraction is positive or negative for x close to 0.

A similar thing happens in the second part of the problem where the limit is as x is going to 2.