Limits by Substitution | Math Coach Amy

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\(\lim_{x \to a} f(x) = f(a)\)

when the expression is defined at a

What to Remember

Plug in when it is allowed

If the expression is defined at the x-value, the limit is the value of the function there.

\(\lim_{x \to a} f(x) = f(a)\) when f is a polynomial, a rational with nonzero denominator, or another continuous expression at a.

Substitute, then simplify

Write the substituted expression on its own line. Then simplify. Do not chain the original limit, the substitution, and the answer with extra equals signs on one line.

Watch the denominator

If substituting makes the denominator 0, substitution is not the whole story. Factor, cancel, or use another method.

Rational exponents

Rewrite \(u^{\frac{2}{3}}\) as \(\bigl(u^{\frac{1}{3}}\bigr)^2\). Take the cube root first, then square.

Level 1

Easy

Polynomials. Substitute, then simplify.

Example

Find \(\lim_{x \to 3} 2x(x + 4).\)

Substitute \(x = 3\).

\(2(3)(3 + 4)\)

\(2(3)(7)\)

\(6 \cdot 7\)

\(42\)

Answer: \(42\)

Find \(\lim_{x \to 4} x^2 - 5x + 1.\)

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Substitute \(x = 4\).

\(4^2 - 5(4) + 1\)

\(16 - 20 + 1\)

\(-3\)

Answer: \(-3\)

Find \(\lim_{x \to -1} 3x^2(x - 2).\)

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Substitute \(x = -1\).

\(3(-1)^2(-1 - 2)\)

\(3(1)(-3)\)

\(-9\)

Answer: \(-9\)

Find \(\lim_{x \to 2} 4x - 7.\)

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Substitute \(x = 2\).

\(4(2) - 7\)

\(8 - 7\)

\(1\)

Answer: \(1\)

Find \(\lim_{x \to 0} (x + 1)(x - 5).\)

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Substitute \(x = 0\).

\((0 + 1)(0 - 5)\)

\((1)(-5)\)

\(-5\)

Answer: \(-5\)

Find \(\lim_{x \to -2} x^3 - 2x.\)

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Substitute \(x = -2\).

\((-2)^3 - 2(-2)\)

\(-8 + 4\)

\(-4\)

Answer: \(-4\)

Level 2

Medium

Rationals that are defined at the x-value.

Example

Find \(\lim_{x \to -1} \dfrac{x^2 + 1}{x + 4}.\)

Substitute \(x = -1\).

\(\dfrac{(-1)^2 + 1}{-1 + 4}\)

\(\dfrac{1 + 1}{3}\)

\(\dfrac{2}{3}\)

Answer: \(\dfrac{2}{3}\)

Find \(\lim_{x \to 3} \dfrac{x^2 - 2x}{x + 5}.\)

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Substitute \(x = 3\).

\(\dfrac{3^2 - 2(3)}{3 + 5}\)

\(\dfrac{9 - 6}{8}\)

\(\dfrac{3}{8}\)

Answer: \(\dfrac{3}{8}\)

Find \(\lim_{x \to 0} \dfrac{2x + 1}{x^2 + 4}.\)

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Substitute \(x = 0\).

\(\dfrac{2(0) + 1}{0 + 4}\)

\(\dfrac{1}{4}\)

Answer: \(\dfrac{1}{4}\)

Find \(\lim_{x \to -1} \dfrac{x^3 + 1}{x - 2}.\)

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Substitute \(x = -1\).

\(\dfrac{(-1)^3 + 1}{-1 - 2}\)

\(\dfrac{-1 + 1}{-3}\)

\(\dfrac{0}{-3}\)

\(0\)

Answer: \(0\)

Find \(\lim_{x \to 2} \dfrac{4x}{x^2 - 9}.\)

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Substitute \(x = 2\).

\(\dfrac{4(2)}{2^2 - 9}\)

\(\dfrac{8}{4 - 9}\)

\(\dfrac{8}{-5}\)

\(-\dfrac{8}{5}\)

Answer: \(-\dfrac{8}{5}\)

Find \(\lim_{x \to 1} (2x - 1)^3.\)

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Substitute \(x = 1\).

\((2(1) - 1)^3\)

\((1)^3\)

\(1\)

Answer: \(1\)

Level 3

Hard

A rational exponent, or a fraction that still simplifies after substituting.

Example

Find \(\lim_{x \to 7} (x - 15)^{\frac{2}{3}}.\)

\(u^{\frac{2}{3}} = \bigl(u^{\frac{1}{3}}\bigr)^2\)

Substitute \(x = 7\).

\((7 - 15)^{\frac{2}{3}}\)

\((-8)^{\frac{2}{3}}\)

\(\bigl((-8)^{\frac{1}{3}}\bigr)^2\)

\((-2)^2\)

\(4\)

Answer: \(4\)

Find \(\lim_{x \to -1} (x + 9)^{\frac{2}{3}}.\)

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\(u^{\frac{2}{3}} = \bigl(u^{\frac{1}{3}}\bigr)^2\)

Substitute \(x = -1\).

\((-1 + 9)^{\frac{2}{3}}\)

\(8^{\frac{2}{3}}\)

\(\bigl(8^{\frac{1}{3}}\bigr)^2\)

\(2^2\)

\(4\)

Answer: \(4\)

Find \(\lim_{x \to 1} (2x - 10)^{\frac{2}{3}}.\)

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Substitute \(x = 1\).

\((2 - 10)^{\frac{2}{3}}\)

\((-8)^{\frac{2}{3}}\)

\(\bigl((-8)^{\frac{1}{3}}\bigr)^2\)

\((-2)^2\)

\(4\)

Answer: \(4\)

Find \(\lim_{x \to 4} \dfrac{x^2 - 16}{x + 3}.\)

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The denominator is not 0, so substitute \(x = 4\).

\(\dfrac{4^2 - 16}{4 + 3}\)

\(\dfrac{16 - 16}{7}\)

\(0\)

Answer: \(0\)

Find \(\lim_{x \to -2} \dfrac{(x + 2)^2}{x^2 + 1}.\)

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Substitute \(x = -2\).

\(\dfrac{(-2 + 2)^2}{(-2)^2 + 1}\)

\(\dfrac{0}{4 + 1}\)

\(0\)

Answer: \(0\)

Find \(\lim_{x \to 4} x^{\frac{3}{2}}.\)

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\(x^{\frac{3}{2}} = \bigl(x^{\frac{1}{2}}\bigr)^3\)

Substitute \(x = 4\).

\(\bigl(4^{\frac{1}{2}}\bigr)^3\)

\(2^3\)

\(8\)

Answer: \(8\)

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