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1.5 Determining Limits Using Algebraic Properties of Limits

Part of Limits and Continuity in AP® Calculus BC. Read the opening section, then create a free account for the full interactive lesson, practice, and progress tracking.

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The algebraic properties of limits allow us to evaluate many limits efficiently. These properties say that the limit of a sum is the sum of the limits, the limit of a difference is the difference of the limits, the limit of a product is the product of the limits, and the limit of a quotient is the quotient of the limits as long as the denominator’s limit is not zero. In many cases, these rules make direct substitution possible because the function is continuous at the point.

These properties are useful because they let us break a complicated expression into smaller pieces. Instead of treating the whole function as one object, we evaluate the behavior of each part separately and then combine the results. This works especially well for polynomials, rational functions with nonzero denominator, roots, exponentials, logarithms in their domains, and many trigonometric functions.

Example 1: Find

\lim_{x \to 3}(2x^2-5x+4)

Since this is a polynomial, we can use the algebraic properties of limits and direct substitution:

\lim_{x \to 3}(2x^2-5x+4) = 2\lim_{x \to 3}(x^2) - 5\lim_{x \to 3}(x) + \lim_{x \to 3}(4)

2(3^2) - 5(3) + 4 = 7

\lim_{x \to 3}(2x^2-5x+4)=7

Example 2: Find

\lim_{x \to 2}\frac{x^2+1}{x+3}

Because the denominator does not approach zero, the quotient law applies:

\lim_{x \to 2}\frac{x^2+1}{x+3} = \frac{\lim_{x \to 2}(x^2+1)}{\lim_{x \to 2}(x+3)}

\frac{2^2+1}{2+3}=1

So,

\lim_{x \to 2}\frac{x^2+1}{x+3}=1

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