Unit 1 · Topic 1.5

1.5 Determining Limits Using Algebraic Properties of Limits

Limits follow simple algebra rules: the limit of a sum is the sum of the limits, and the same goes for differences, products, constant multiples and (with a nonzero bottom) quotients. These rules explain why you can often just plug in x = c, which is called direct substitution.

Key terms

limit properties
direct substitution
sum and product of limits
quotient of limits
composite function

The limit properties

Suppose lim (x→c) f(x) = A and lim (x→c) g(x) = B, where A and B are real numbers. Then:

Sum and difference: lim (x→c) [f(x) ± g(x)] = A ± B.
Constant multiple: lim (x→c) k·f(x) = k·A for any constant k.
Product: lim (x→c) f(x)·g(x) = A·B.
Quotient: lim (x→c) f(x) / g(x) = A / B, but only if B ≠ 0.
Powers and roots: lim (x→c) [f(x)]ⁿ = Aⁿ, and lim (x→c) ⁿ√(f(x)) = ⁿ√A when that root is a real number.

Direct substitution

Using the properties step by step, you can show that for any polynomial p, lim (x→c) p(x) = p(c). The same holds for rational functions where the denominator isn't 0 at c, and for sin x, cos x, eˣ, ln x (for x > 0) and roots on their domains. These are all continuous functions, so you can just plug in.

Here is why it works for a polynomial. lim (x→2) (3x² − 5x + 1) splits by the sum and constant multiple rules into 3·lim x² − 5·lim x + lim 1. By the power rule for limits, that's 3(2)² − 5(2) + 1 = 3, which is exactly what you get by plugging in x = 2.

Example: lim (x→π) (cos x + x²) = cos π + π² = π² − 1.

Limits of composite functions

If lim (x→c) g(x) = L and f is continuous at L, then lim (x→c) f(g(x)) = f(L). In words, you can pass a limit inside a continuous outer function.

So lim (x→2) √(x² + 5) = √(lim (x→2) (x² + 5)) = √9 = 3. If the outer function is not continuous at L, you need more care; that case shows up in 1.9.

When a property doesn't apply

The properties need the individual limits to exist as real numbers. If the bottom of a quotient approaches 0, the quotient property says nothing. Then: if the top also approaches 0, you have the indeterminate form 0/0 and need to rewrite (1.6). If the top approaches a nonzero number, the quotient grows without bound, so the limit is infinite or does not exist (1.14).

Worked examples

Try each one yourself first, then open the solution.

Example 1
Combining given limits
lim (x→4) f(x) = 3 and lim (x→4) g(x) = −2. Find (a) lim (x→4) [2f(x) − g(x)]², (b) lim (x→4) √(f(x) + 6), (c) lim (x→4) f(x)·g(x) / (x − 1).
Show the solutionHide the solution
1. Step 1: (a) Inside first: lim [2f(x) − g(x)] = 2(3) − (−2) = 8. Then square: 8² = 64.
2. Step 2: (b) lim (f(x) + 6) = 9, and √ is continuous at 9, so the limit is √9 = 3.
3. Step 3: (c) The top approaches 3(−2) = −6. The bottom approaches 4 − 1 = 3, which is not 0, so the quotient property applies: −6/3 = −2.
Answer: (a) 64, (b) 3, (c) −2
Example 2
Trap: a zero denominator
lim (x→4) f(x) = 3 and lim (x→4) h(x) = 0. A student writes lim (x→4) f(x)/h(x) = 3/0 = 0. What's wrong, and what can you conclude?
Show the solutionHide the solution
1. Step 1: The quotient property requires the limit of the bottom to be nonzero. Here it is 0, so the property can't be used, and 3/0 is not a number at all, let alone 0.
2. Step 2: The top heads toward 3 while the bottom gets tiny. Dividing something near 3 by something near 0 gives outputs that grow huge in size.
3. Step 3: So the outputs don't approach any real number. Depending on the signs of h(x) near 4, the limit is ∞, −∞, or doesn't exist in any sense (if the signs differ on each side).
Answer: The limit is not 0. Since the top approaches 3 and the bottom approaches 0, the quotient is unbounded near x = 4, so the limit is not a finite number.

Common mistakes

Using the quotient property when the bottom's limit is 0.
Plugging in when the function isn't continuous at c, like at the break in a piecewise function. Check one-sided limits there instead.
Assuming lim f(g(x)) = f(lim g(x)) when f has a jump or hole at that value.

On the exam

Multiple-choice questions often give limits of f and g (or their graphs) and ask for the limit of a combination. Find each piece, then combine.
If direct substitution gives a real number with a nonzero denominator, that's your answer. Don't overcomplicate it.

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