Math Problem Statement

Graph this a) 

f

(

2

)

=

0

f(−2)=0

b) 

lim

x

2

f

(

x

)

=

4

lim 

x→−2 

−  

 f(x)=4

c) 

lim

x

2

f

(

x

)

=

0

lim 

x→−2 

 

 f(x)=0

d) 

lim

x

2

f

(

x

)

lim 

x→−2

 f(x) does not exist.

e) 

f

(

4

)

=

4

f(−4)=4

f) 

lim

x

4

f

(

x

)

=

4

lim 

x→−4 

−  

 f(x)=4

g) 

lim

x

4

f

(

x

)

=

4

lim 

x→−4 

 

 f(x)=4

h) 

lim

x

4

f

(

x

)

=

4

lim 

x→−4

 f(x)=4

Solution

Here is the graph of f(x)f(x) based on the conditions provided:

  • f(2)=0f(-2) = 0.
  • limx2f(x)=4\lim_{x \to -2^-} f(x) = 4, meaning as xx approaches 2-2 from the left, f(x)f(x) approaches 4.
  • limx2+f(x)=0\lim_{x \to -2^+} f(x) = 0, meaning as xx approaches 2-2 from the right, f(x)f(x) approaches 0.
  • limx2f(x)\lim_{x \to -2} f(x) does not exist because the left and right limits are different.
  • f(4)=4f(-4) = 4.
  • limx4f(x)=4\lim_{x \to -4^-} f(x) = 4, meaning as xx approaches 4-4 from the left, f(x)f(x) remains 4.
  • limx4+f(x)=4\lim_{x \to -4^+} f(x) = 4, meaning as xx approaches 4-4 from the right, f(x)f(x) also remains 4.
  • limx4f(x)=4\lim_{x \to -4} f(x) = 4, so the limit at 4-4 exists.

Would you like more details on the graph or any of the limit concepts?

Here are 5 related questions to consider:

  1. How do left-hand and right-hand limits affect the overall limit at a point?
  2. What happens if the left-hand and right-hand limits do not match?
  3. Can you find similar examples where a limit does not exist at certain points?
  4. How do asymptotes relate to discontinuities in functions?
  5. What is the role of piecewise functions in modeling real-world problems?

Tip: When dealing with limits, always check both the left-hand and right-hand behavior to determine if the overall limit exists.

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Math Problem Analysis

Mathematical Concepts

Limits
Discontinuities
Piecewise Functions

Formulas

Limit definition: lim x→a f(x)
Left-hand limit: lim x→a− f(x)
Right-hand limit: lim x→a+ f(x)

Theorems

Limit Theorem
Existence of Limits Theorem

Suitable Grade Level

Grades 11-12 (Calculus)