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Math Coach Amy
Amy Ferguson Moncure

Calculus BC · Unit 1

Piecewise Limits

Use the left piece from the left and the right piece from the right. The value at the break does not decide the limit.

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Key Ideas Easy Medium Hard

\(x \to a^-\) left piece

\(x \to a^+\) right piece

What to Remember

One side, one piece

The left-hand limit uses the piece for \(x < a\). The right-hand limit uses the piece for \(x > a\).

The value at a does not matter

\(f(a)\) can be a different number. The two-sided limit only cares whether the one-sided limits match.

Absolute value splits

\(|x - a| = a - x\) on the left and \(x - a\) on the right. Rewrite each side, then take the limit.

Both sides equal

If the one-sided limits are equal, that common number is the two-sided limit — even if \(f\) is defined as something else at \(a\).

Level 1

Easy

Read the correct piece, then substitute.

Example
Let \(f(x) = \begin{cases} 3x - 1 & x < 2 \\ 5 & x = 2 \\ x + 4 & x > 2 \end{cases}\). Find \(\lim_{x \to 2^+} f(x)\).

From the right, use \(x + 4\).

Substitute \(x = 2\).

\(2 + 4\)

\(6\)

Answer: \(6\)

1
Let \(f(x) = \begin{cases} 3x - 1 & x < 2 \\ 5 & x = 2 \\ x + 4 & x > 2 \end{cases}\). Find \(\lim_{x \to 2^-} f(x)\).
Show Answer

From the left, use \(3x - 1\).

Substitute \(x = 2\).

\(3(2) - 1\)

\(5\)

Answer: \(5\)

2
Let \(g(x) = \begin{cases} x^2 & x \leq 1 \\ 4x - 1 & x > 1 \end{cases}\). Find \(\lim_{x \to 1^+} g(x)\).
Show Answer

From the right, use \(4x - 1\).

\(4(1) - 1\)

\(3\)

Answer: \(3\)

3
Let \(g(x) = \begin{cases} x^2 & x \leq 1 \\ 4x - 1 & x > 1 \end{cases}\). Find \(\lim_{x \to 1^-} g(x)\).
Show Answer

From the left, use \(x^2\).

\(1^2\)

\(1\)

Answer: \(1\)

4
Let \(h(x) = \begin{cases} 2x + 3 & x < 0 \\ 1 & x = 0 \\ 3 - x & x > 0 \end{cases}\). Find \(\lim_{x \to 0^-} h(x)\).
Show Answer

From the left, use \(2x + 3\).

\(2(0) + 3\)

\(3\)

Answer: \(3\)

5
Let \(h(x) = \begin{cases} 2x + 3 & x < 0 \\ 1 & x = 0 \\ 3 - x & x > 0 \end{cases}\). Find \(h(0)\).
Show Answer

The middle piece says \(h(0) = 1\).

\(1\)

Answer: \(1\)

Level 2

Medium

Decide whether the two-sided limit exists.

Example
Let \(f(x) = \begin{cases} 3x - 1 & x < 2 \\ 5 & x = 2 \\ x + 4 & x > 2 \end{cases}\). Find \(\lim_{x \to 2} f(x)\).

\(\lim_{x \to 2^-} f(x) = 5\)

\(\lim_{x \to 2^+} f(x) = 6\)

The one-sided limits are not equal.

does not exist

Answer: does not exist

1
Let \(f(x) = \begin{cases} 4x - 2 & x < 3 \\ 7 & x = 3 \\ 2x + 4 & x > 3 \end{cases}\). Find \(\lim_{x \to 3} f(x)\).
Show Answer

Left: \(4(3) - 2 = 10\).

Right: \(2(3) + 4 = 10\).

The one-sided limits both equal 10.

\(10\)

Answer: \(10\)

2
Let \(g(x) = \begin{cases} x^2 - 1 & x < 1 \\ 4 & x = 1 \\ 2x - 2 & x > 1 \end{cases}\). Find \(\lim_{x \to 1} g(x)\).
Show Answer

Left: \(1 - 1 = 0\).

Right: \(2 - 2 = 0\).

\(0\)

Answer: \(0\)

3
Let \(k(x) = \begin{cases} -x + 5 & x \leq -2 \\ x^2 + 1 & x > -2 \end{cases}\). Find \(\lim_{x \to -2} k(x)\).
Show Answer

Left: \(-(-2) + 5 = 7\).

Right: \(4 + 1 = 5\).

does not exist

Answer: does not exist

4
Let \(p(x) = \begin{cases} 2x + 1 & x < 4 \\ x - 3 & x \geq 4 \end{cases}\). Find \(\lim_{x \to 4} p(x)\).
Show Answer

Left: \(8 + 1 = 9\).

Right: \(4 - 3 = 1\).

does not exist

Answer: does not exist

5
Let \(q(x) = \begin{cases} 5 - x & x < 5 \\ 0 & x = 5 \\ x - 5 & x > 5 \end{cases}\). Find \(\lim_{x \to 5} q(x)\).
Show Answer

Left: \(5 - 5 = 0\).

Right: \(5 - 5 = 0\).

\(0\)

Answer: \(0\)

Level 3

Hard

Absolute value in the denominator, so the two sides look different.

Example
Let \(f(x) = \begin{cases} \dfrac{(x + 2)^2(x - 1)}{|x + 2|} & x \neq -2 \\ 4 & x = -2 \end{cases}\). Find \(\lim_{x \to -2} f(x)\).

\(|x + 2| = -(x + 2)\) when \(x < -2\), and \(x + 2\) when \(x > -2\).

\(\dfrac{(x + 2)^2}{|x + 2|} = |x + 2|\)

So \(f(x) = |x + 2|(x - 1)\) for \(x \neq -2\).

Left: \(\lim_{x \to -2^-} |x + 2|(x - 1) = 0 \cdot (-3) = 0\)

Right: \(\lim_{x \to -2^+} |x + 2|(x - 1) = 0 \cdot (-3) = 0\)

\(0\)

Answer: \(0\)

1
Let \(f(x) = \begin{cases} \dfrac{(x - 4)^2(x + 1)}{|x - 4|} & x \neq 4 \\ 3 & x = 4 \end{cases}\). Find \(\lim_{x \to 4} f(x)\).
Show Answer

\(f(x) = |x - 4|(x + 1)\) for \(x \neq 4\).

\(\lim_{x \to 4} |x - 4|(x + 1) = 0 \cdot 5 = 0\)

\(0\)

Answer: \(0\)

2
Find \(\lim_{x \to 0} \dfrac{|x|}{x}.\)
Show Answer

Left: \(|x| = -x\), so \(\dfrac{-x}{x} = -1\).

Right: \(|x| = x\), so \(\dfrac{x}{x} = 1\).

does not exist

Answer: does not exist

3
Find \(\lim_{x \to -3} \dfrac{|x + 3|}{x + 3}.\)
Show Answer

This is \(\dfrac{|u|}{u}\) with \(u = x + 3\).

Left-hand limit is \(-1\).

Right-hand limit is \(1\).

does not exist

Answer: does not exist

4
Find \(\lim_{x \to 1} \dfrac{(x - 1)|x - 1|}{x - 1}.\)
Show Answer

\(\dfrac{(x - 1)|x - 1|}{x - 1} = |x - 1|\) for \(x \neq 1\).

\(\lim_{x \to 1} |x - 1| = 0\)

\(0\)

Answer: \(0\)

5
Let \(f(x) = \begin{cases} \dfrac{(x - 2)^2(x + 5)}{|x - 2|} & x \neq 2 \\ 7 & x = 2 \end{cases}\). Find \(\lim_{x \to 2} f(x)\).
Show Answer

\(f(x) = |x - 2|(x + 5)\) for \(x \neq 2\).

\(\lim_{x \to 2} |x - 2|(x + 5) = 0\)

\(0\)

Answer: \(0\)

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