# Left and Right Hand limit of a Piece wise function?

1. Aug 10, 2013

### kashan123999

1. The problem statement, all variables and given/known data

Find the left hand limit and right hand limit for x--> 2 for the function

f(x) = 2x+1 if 0 ≤ x ≤ 2
7-x if 2 ≤ x ≤ 4
x if 4 ≤ x ≤ 6
2. Relevant equations

Lim f(x) for x->2 = ?

3. The attempt at a solution

I honestly don't know where to start please help me,ok we can answer the question by plugging 2 in first rule and finding LEFT hand limit,and plugging 2 in the 2nd rule and finding right hand limit...but i just couldn't find the logic behind this all method and also why didn't we use the third rule? please explain in layman terms as english is not my native language

2. Aug 10, 2013

### LCKurtz

To answer your last question first, you don't use the third rule because you are interested in values of $x$ near $2$ and the third rule is irrelevant to that.

You don't evaluate limits by plugging in the value unless you know the functions you are using are continuous. But in your case they are both polynomials. Assuming you have the theorem for polynomials $p(x)$ that$$\lim_{x \rightarrow a}p(x) = p(a)$$then, for example, your left hand limit is$$\lim_{x\rightarrow 2^-}2x+1 =2\cdot 2 + 1=5$$Similarly for the right hand limit. If they come out equal, that would give the two sided limit, otherwise it doesn't exist.

3. Aug 10, 2013

### kashan123999

sorry i couldn't understand that "irrelevant thing"...

4. Aug 11, 2013

### LCKurtz

"Irrelevant" means it has nothing to do with your problem. Ignore the third rule since it is for $x>4$, not $x$ near $2$.

5. Aug 11, 2013

### kashan123999

ahan Thank you very much sir