a) $lim_{x\rightarrow3}f(x)=$\nb) $f(3)=$\nc) $lim_{x\rightarrow1}f(x)=$\nd) $f(1)=$\ne) $lim_{x\rightarrow2}…

a) $lim_{x\rightarrow3}f(x)=$\nb) $f(3)=$\nc) $lim_{x\rightarrow1}f(x)=$\nd) $f(1)=$\ne) $lim_{x\rightarrow2}f(x)=$\nf) $lim_{x\rightarrow - 2^{+}}f(x)=$\ng) $lim_{x\rightarrow - 2^{-}}f(x)=$\nh) $f(-2)=$

a) $lim_{x\rightarrow3}f(x)=$\nb) $f(3)=$\nc) $lim_{x\rightarrow1}f(x)=$\nd) $f(1)=$\ne) $lim_{x\rightarrow2}f(x)=$\nf) $lim_{x\rightarrow - 2^{+}}f(x)=$\ng) $lim_{x\rightarrow - 2^{-}}f(x)=$\nh) $f(-2)=$

Answer

Explanation:

Step1: Analyze limit as x approaches 3

As (x) approaches 3 from both the left - hand and right - hand sides, the (y) - value of the function approaches 2. So, (\lim_{x\rightarrow3}f(x)=2).

Step2: Find the value of (f(3))

The function is not defined at (x = 3) (open - circle at (x = 3)), so (f(3)) is undefined.

Step3: Analyze limit as x approaches 1

As (x) approaches 1 from both the left - hand and right - hand sides, the (y) - value of the function approaches 4. So, (\lim_{x\rightarrow1}f(x)=4).

Step4: Find the value of (f(1))

The filled - in circle at (x = 1) has a (y) - value of 2, so (f(1)=2).

Step5: Analyze limit as x approaches 2

As (x) approaches 2 from both the left - hand and right - hand sides, the (y) - value of the function approaches 2. So, (\lim_{x\rightarrow2}f(x)=2).

Step6: Analyze right - hand limit as x approaches - 2

As (x) approaches (-2) from the right - hand side ((x\rightarrow - 2^{+})), the (y) - value of the function approaches 2. So, (\lim_{x\rightarrow - 2^{+}}f(x)=2).

Step7: Analyze left - hand limit as x approaches - 2

As (x) approaches (-2) from the left - hand side ((x\rightarrow - 2^{-})), the (y) - value of the function approaches 2. So, (\lim_{x\rightarrow - 2^{-}}f(x)=2).

Step8: Find the value of (f(-2))

The filled - in circle at (x=-2) has a (y) - value of (-2), so (f(-2)=-2).

Answer:

a) (\lim_{x\rightarrow3}f(x)=2) b) (f(3)) is undefined c) (\lim_{x\rightarrow1}f(x)=4) d) (f(1)=2) e) (\lim_{x\rightarrow2}f(x)=2) f) (\lim_{x\rightarrow - 2^{+}}f(x)=2) g) (\lim_{x\rightarrow - 2^{-}}f(x)=2) h) (f(-2)=-2)