Limits f = x - 2 for x <= 3, x - 1 for x > 3

[market]

Find lim f(x)
x approaches 3

Does it not exist b/c it doesn't state if the 3 is approaches from the left or right? So there is no way to plot a line on the graph? I would have thought the answer was 1. 3-2=1 but I guess I was wrong Please explain.
Thanks

 

[pka]

As a shorthand notation use \(\displaystyle f(a + ) = \lim _{x \to a^ + } (x)\), that is the right-hand limit at \(\displaystyle x=a\) and for the left-hand limit \(\displaystyle f(a - ) = \lim _{x \to a^ - } (x)\).

In order for a limit to exits at all we must have \(\displaystyle f(a - ) = f(a + ) = \lim _{x \to a} (x)\). That is, we must have a left-hand limit, a right-hand limit and those two must be equal.

Does that happen in this case?

 

[market]

I would say yes b/c I thought the answer would be 1 either way. The teacher has not thought us anything so I don't know exactly what to do. I wouldn't know how to graph these. left or right wither way 3-2=1 but I guess I am not seeing something

I'm confused.

 

[pka]

\(\displaystyle f(3+)=2\) & \(\displaystyle f(3-)=1\).
Think about why those are true.
Can you explain why?

 

[market]

Re: Limits f = x - 2 for, x <= 3 x - 1 for x > 3

I used x - 2 for, x <= 3 I don't understand why there is a different answer for f(x-) vs f(x+) why would 3+ be 2? I don't know why it's different.

 

[pka]

Look very closely at the function.
For numbers on the left of 3 \(\displaystyle x < 3\) we subtract 2 from them.
Therefore, the left-hand limit is 1

For numbers on the right of 3 \(\displaystyle x > 3\) we subtract 1 from them.
Therefore, the left-hand limit is 2.

For example, 2.99 is on the left of 3, so \(\displaystyle f(2.99)=.99\) almost 1.
For example, 3.01 is on the right of 3, so \(\displaystyle f(3.01)=2.01\) almost 2.

 

[market]

Wow I think I learn something new everything you help me with an answer. I am realizing that I do not know anything thanks to my horrible teacher.

 

[pka]

I think I learn something new everything you help me with an answer. I am realizing that I do not know anything

Well, I have taught some form of calculus from 1964 to 2005.
At some point in there (1972) is wrote a successful PhD thesis.
In the 1980’s I became an ardent supporter of Calculus Reform.
I do not know what you level is.
So what I am about to say will offend some and I do not care.
If any secondary school offers a course in The Calculus, the instructor must have a graduate degree in mathematics and must be AP certified.
Unfortunately, that is almost never the case! Yours may be such a case!
The drive for prestige overrides any true educational concerns.
Good looks overrides TRUTH.