Left and Right Hand limit of a Piece wise function?

[kashan123999]

Homework Statement

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

Homework Equations

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

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

 

[LCKurtz]

Homework Statement

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

Homework Equations

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

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

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.

 

[kashan123999]

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.

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

 

[LCKurtz]

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

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

 

[kashan123999]

ahan Thank you very much sir