This function has a jump discontinuity at x = 1. The left-hand limit is lim(x→1⁻) 2 = 2, the function value is h(1) = 2, and the right-hand limit is lim(x→1⁺) 5 = 5. Since the left and right limits exist but are unequal (2 ≠ 5), this creates a jump discontinuity. The function jumps from 2 on the left to 5 on the right at x = 1. Students might think this is continuous because h(1) equals the left-hand limit, but continuity requires both one-sided limits to be equal. To identify jump discontinuities: verify that both one-sided limits exist but have different values.

For the piecewise function g(x), we need to check the left and right limits at x=1. From the left: lim(x→1⁻) g(x) = 1+2 = 3. From the right: lim(x→1⁺) g(x) = 5-1 = 4. Since the left limit (3) doesn't equal the right limit (4), the overall limit doesn't exist at x=1. When one-sided limits exist but are unequal, this creates a jump discontinuity. Students often confuse this with a removable discontinuity, but removable discontinuities require the two-sided limit to exist. The key strategy: for piecewise functions, always check both one-sided limits at transition points—if they're finite but different, it's a jump discontinuity.

The function $q(x)=\frac{|x|}{x}$ has a jump discontinuity at $x=0$. For $x<0$, we have $|x|=-x$, so $q(x)=\frac{-x}{x}=-1$. For $x>0$, we have $|x|=x$, so $q(x)=\frac{x}{x}=1$. Therefore, $\lim_{x\to 0^-}q(x)=-1$ and $\lim_{x\to 0^+}q(x)=1$. The function is undefined at $x=0$ due to division by zero. Since the one-sided limits exist but are different (-1 vs 1), this is a jump discontinuity. Students might think this is infinite because of division by zero, but the absolute value in the numerator ensures finite one-sided limits. A jump discontinuity is characterized by finite but unequal one-sided limits.

The function s(x) = |x|/x has a jump discontinuity at x = 0. For x > 0, |x| = x, so s(x) = x/x = 1. For x < 0, |x| = -x, so s(x) = -x/x = -1. Therefore, the right-hand limit is 1 and the left-hand limit is -1. Since these one-sided limits are different, this is a jump discontinuity. Note that s(0) is undefined (0/0), but this doesn't affect the classification—the differing one-sided limits determine it's a jump. Students might think this is an infinite discontinuity because of the division by zero, but the absolute value in the numerator prevents infinite behavior. To identify discontinuities with absolute values: evaluate the one-sided limits carefully, considering how the absolute value behaves on each side.

The function $s(x)=\frac{x+1}{x^2-1}$ has an infinite discontinuity at $x=1$. We can factor the denominator as $(x-1)(x+1)$, giving $s(x)=\frac{x+1}{(x-1)(x+1)}=\frac{1}{x-1}$ for $x\neq -1$. As $x$ approaches 1 from the left, $x-1$ approaches 0 through negative values, so $\lim_{x\to 1^-}s(x)=-\infty$. As $x$ approaches 1 from the right, $x-1$ approaches 0 through positive values, so $\lim_{x\to 1^+}s(x)=+\infty$. Students might think this is removable because the $(x+1)$ factor appears to cancel, but the cancellation only occurs when $x\neq -1$, not at $x=1$. An infinite discontinuity occurs when at least one one-sided limit is infinite, creating a vertical asymptote at the point of discontinuity.

The function $v(x)=\frac{x-4}{|x-4|}$ has a jump discontinuity at $x=4$. For $x<4$, we have $x-4<0$, so $|x-4|=-(x-4)$, giving $v(x)=\frac{x-4}{-(x-4)}=-1$. For $x>4$, we have $x-4>0$, so $|x-4|=x-4$, giving $v(x)=\frac{x-4}{x-4}=1$. Therefore, $\lim_{x\to 4^-}v(x)=-1$ and $\lim_{x\to 4^+}v(x)=1$. The function is undefined at $x=4$ due to division by zero. Since the one-sided limits exist but differ (-1 vs 1), this is a jump discontinuity. This function is similar to $\frac{|x|}{x}$ but shifted to $x=4$. Jump discontinuities are characterized by finite but unequal one-sided limits, creating a "jump" in the function's graph.

This function has no discontinuity at x = 2. The left-hand limit is lim(x→2⁻) 3x = 6, the right-hand limit is lim(x→2⁺) 6 = 6, and Z(2) = 6. Since both one-sided limits exist, are equal, and match the function value, the function is continuous at x = 2. This is an example of a piecewise function that is carefully constructed to be continuous at the boundary point. Students might assume piecewise functions are automatically discontinuous, but continuity depends on matching limits and function values. To verify continuity: check that both one-sided limits exist, are equal, and match the function value.

This function has a removable discontinuity at x = 0. The expression n(x) = (x²+3x)/x = x(x+3)/x simplifies to n(x) = x+3 for all x ≠ 0, but n(0) is undefined since we get 0/0. The limit as x approaches 0 is lim(x→0) (x+3) = 3, but the function is undefined at x = 0. This creates a removable discontinuity because we could define n(0) = 3 to make the function continuous. Students often assume division by zero creates infinite discontinuities, but when both numerator and denominator approach zero, check for simplification. To identify removable discontinuities: look for 0/0 forms that resolve to finite limits after factoring.

This function has a removable discontinuity at x = -4. The expression r(x) = (x+4)/(x+4) simplifies to r(x) = 1 for all x ≠ -4, but r(-4) is undefined since we get 0/0. The limit as x approaches -4 is lim(x→-4) 1 = 1, but the function is undefined at x = -4. This creates a removable discontinuity because we could define r(-4) = 1 to make the function continuous. Students often think this is an infinite discontinuity because of the rational form, but the key is that both numerator and denominator approach zero simultaneously. To identify removable discontinuities: look for 0/0 indeterminate forms that can be simplified to find a finite limit.

This function has an infinite discontinuity at x = -1. As x approaches -1, the denominator (x+1)² approaches 0 while the numerator remains 2, causing the function values to approach +∞ from both sides. Since (x+1)² is always positive, we get lim(x→-1) 2/(x+1)² = +∞. This creates a vertical asymptote at x = -1, which defines an infinite discontinuity. Students sometimes confuse this with a removable discontinuity, but the key is that the numerator doesn't approach zero as the denominator does. To identify infinite discontinuities: look for points where the denominator approaches zero while the numerator approaches a non-zero value.