Calculus - Limits Review

Evaluate

 $\lim_{x\rightarrow1^+}f\left(x\right)$  

• DNE

• 1

• 4

• -2

Evaluate

 $\lim_{x\rightarrow1^-}f\left(x\right)$  

• DNE

• 1

• 4

• -2

Tags

CCSS.HSF.IF.B.4

CCSS.HSF.IF.C.7

Evaluate

 $\lim_{x\rightarrow1^{ }}f\left(x\right)$  

• 2

• 4

• DNE

• -2

Tags

CCSS.HSF.IF.A.2

Evaluate

 $\lim_{x\rightarrow-4}f\left(x\right)$  

• 2

• -4

• DNE

• -2

Tags

CCSS.HSF.IF.A.2

Is the function continuous at x = 2?

• Yes

• No; because f(2) isn't defined

• No; the limit at x=2 doesn't exist

• No; the limit at x=2 doesn't equal f(2)

Tags

CCSS.8.F.A.1

CCSS.HSF.IF.B.5

Is the function continuous at x = 4?

• Yes

• No; because f(4) isn't defined

• No; the limit at x=4 doesn't exist

• No; the limit at x=4 doesn't equal f(4)

Tags

CCSS.HSF-IF.C.7B

What are you finding when you calculate

 $\lim_{x\rightarrow\pm\infty}f\left(x\right)$  ?

• Vertical asymptote(s) of f(x)

• Horizontal asymptote(s) of f(x)

• Values where the limit DNE

• You can't calculate this type of limit

Tags

CCSS.HSF-IF.C.7D

If you get #/0 when evaluating a limit, which of the following does NOT apply?

• you should try to simplify and re-evaluate

• the graph has a vertical asymptote at the value of x

• the limit does not exist

• the limit could be

  $\pm\infty$

If you get 0/0 when evaluating a limit, which of the following is NOT true?

• you could try to simplify and re-evaluate

• you could apply L'Hopital's Rule

• the function most likely has a hole in its graph

• the limit does not exist

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Name the discontinuity at x=2. 

• Infinite Discontinuity
• Removable Discontinuity
• Jump Discontinuity
• Continuous at x=2

Tags

CCSS.HSF-IF.C.7D