Question

1. determine the point(s), if any, at which each function is discontinuous. classify any discontinuity as jump, removable, infinite, or other. f(x) = \begin{cases} 1 + x, & \text{if } x < 0, \\ 2, & \text{if } x = 0, \\ 1 - x^2, & \text{if } 0 < x < 2, \\ 4 - x, & \text{if } x geq 2. end{cases}

Explanation:

Step1: Check continuity at \( x = 0 \)

• Left - hand limit (LHL) as \( x

ightarrow0^{-} \):

• For \( x < 0 \), \( f(x)=1 + x \). So, \(\lim_{x

ightarrow0^{-}}f(x)=\lim_{x
ightarrow0^{-}}(1 + x)\).

• Substitute \( x = 0 \) into \( 1 + x \), we get \( 1+0 = 1 \).
• Right - hand limit (RHL) as \( x

ightarrow0^{+} \):

• For \( 0 < x < 2 \), \( f(x)=1 - x^{2} \). So, \(\lim_{x

ightarrow0^{+}}f(x)=\lim_{x
ightarrow0^{+}}(1 - x^{2})\).

• Substitute \( x = 0 \) into \( 1 - x^{2} \), we get \( 1-0 = 1 \).
• Value of the function at \( x = 0 \): \( f(0)=2 \).
• Since \(\lim_{x

ightarrow0^{-}}f(x)=\lim_{x
ightarrow0^{+}}f(x) = 1\) but \( f(0)=2
eq1 \), the function has a removable discontinuity at \( x = 0 \) (because the limit exists but is not equal to the function value at that point).

Step2: Check continuity at \( x = 2 \)

• Left - hand limit (LHL) as \( x

ightarrow2^{-} \):

• For \( 0 < x < 2 \), \( f(x)=1 - x^{2} \). So, \(\lim_{x

ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{-}}(1 - x^{2})\).

• Substitute \( x = 2 \) into \( 1 - x^{2} \), we get \( 1-2^{2}=1 - 4=-3 \).
• Right - hand limit (RHL) as \( x

ightarrow2^{+} \):

• For \( x\geq2 \), \( f(x)=4 - x \). So, \(\lim_{x

ightarrow2^{+}}f(x)=\lim_{x
ightarrow2^{+}}(4 - x)\).

• Substitute \( x = 2 \) into \( 4 - x \), we get \( 4 - 2 = 2 \).
• Since \(\lim_{x

ightarrow2^{-}}f(x)=-3\) and \(\lim_{x
ightarrow2^{+}}f(x)=2\) and \(-3
eq2\), the function has a jump discontinuity at \( x = 2 \) (because the left - hand limit and right - hand limit exist but are not equal).

Answer:

• At \( x = 0 \): Removable discontinuity (limit is \( 1 \), \( f(0)=2 \)).
• At \( x = 2 \): Jump discontinuity (LHL \(=-3\), RHL \( = 2\)).