Limits

2.3 Basic Limit Laws

In Section 2.1.4 we relied primarily on graphical and numerical approaches to investigate limits and estimate their values. In the next four sections we go beyond this intuitive approach and develop tools for computing limits in a precise way. The next theorem provides our first set of tools.

The proof of Theorem 1 is discussed in Section 2.9 and Appendix D. To illustrate the underlying idea, consider two numbers such as \(2.99\) and \(5.0001\). Observe that \(2.99\) is close to \(3\) and \(5.0001\)is close to \(5\), so certainly the sum \(2.99 + 5.0001\) is close to \(3 + 5\) and the product \((2.99)(5.0001)\) is close to\((3)(5)\). In the same way, if \(f(x)\) approaches \(L\) and \(g(x)\) approaches \(M\) as \(x\rightarrow c\), then \(f(x) + g(x)\) approaches the sum \(L+M\) , and \(f(x)g(x)\) approaches the product \(LM\). The other laws are similar.

THEOREM 1 Basic Limit Laws

Let \(L, M \in \mathbb{R}\). If \(\lim\limits_{x\rightarrow c} f(x) = L\) and \(\lim\limits_{x\rightarrow c} g(x) = M\) exist, then

(i) Sum Law: \(\lim\limits_{x\rightarrow c} (f(x)+g(x))\) exists and \[\lim\limits_{x\rightarrow c} (f(x)+g(x)) = \lim\limits_{x\rightarrow c} f(x) + \lim\limits_{x\rightarrow c} g(x) = L+M\].
(ii) Constant Multiple Law: For any number \(k\), \(\lim\limits_{x\rightarrow c} (kf(x))\) exists and \[\lim\limits_{x\rightarrow c} (kf(x))=k\lim\limits_{x\rightarrow c} f(x) = kL\].
(iii) Product Law: \(\lim\limits_{x\rightarrow c} (f(x)g(x))\) exists and \[\lim\limits_{x\rightarrow c} (f(x)g(x)) = \left(\lim\limits_{x\rightarrow c}f(x)\right)\left(\lim\limits_{x\rightarrow c}g(x)\right) = LM\].
(iv) Quotient Law: If \(M \neq 0\), then \(\lim\limits_{x\rightarrow c}\frac{f(x)}{g(x)}\) exists and \[\lim\limits_{x\rightarrow c}\frac{f(x)}{g(x)} = \frac{\lim\limits_{x\rightarrow c} f(x)}{\lim\limits_{x\rightarrow c} g(x)} = \frac{L}{M}\]
(v) Power Law: \(\displaystyle \lim \limits_{x \rightarrow c} [f(x)]^n\) exists and \[\lim \limits_{x \rightarrow c} [f(x)]^n= \Big(\lim \limits_{x \rightarrow c} f(x)\Big)^n\]
(vi) Root Law: \(\displaystyle \lim \limits_{x \rightarrow c} \sqrt[n]{f(x)}\) exists and \[\lim \limits_{x \rightarrow c} \sqrt[n]{f(x)}= \sqrt[n]{\lim \limits_{x \rightarrow c} f(x)}\] Under the following conditions: If \(n\) is even, then we require that either \(L >0\) or if \(L = 0\), then we require that \(\displaystyle f(x) \geq 0\) for \(x\) sufficiently close to \(c\) (except possibly at \(x=c\) itself).

Before proceeding to the examples, we make some useful remarks.

The Sum and Product Laws are valid for any number of functions. For example,\[\lim\limits_{x\rightarrow c} \left(f_1(x)+f_2(x)+f_3(x)\right) = \lim\limits_{x\rightarrow c} f_1(x) + \lim\limits_{x\rightarrow c} f_2(x) + \lim\limits_{x\rightarrow c} f_3(x)\]
The Sum Law has a counterpart for differences: \[\lim\limits_{x\rightarrow c} (f(x)-g(x)) = \lim\limits_{x\rightarrow c} f(x) - \lim\limits_{x\rightarrow c} g(x)\]This follows from the Sum and Constant Multiple Laws (with \(k=-1\)):\[\lim\limits_{x\rightarrow c} (f(x)-g(x)) = \lim\limits_{x\rightarrow c} f(x) + \lim\limits_{x\rightarrow c} (-g(x)) = \lim\limits_{x\rightarrow c} f(x) - \lim\limits_{x\rightarrow c} g(x)\]
Recall two basic limits from Theorem 1 in Section 2.1.4: \(\lim\limits_{x\rightarrow c} k = k\), \(\lim\limits_{x\rightarrow c} x = c\). Applying Law (v) to \(f(x)=x\), we obtain\[\lim\limits_{x\rightarrow c} x^{\tfrac{p}{q}} = c^{\tfrac{p}{q}}\]for integers \(p,q\). Assume that \(c\geq 0\) if \(q\) is even and that \(c > 0\) if \(\tfrac{p}{q} < 0\).
The limit laws also apply to one-sided limits. In particular, if \(c=0, \tfrac{p}{q} >0 \) and \(q\) is even, we get that \[\lim_{x\rightarrow 0^+} x^\frac{p}{q} = 0^\frac{p}{q}=0\], and, for instance, \[\lim_{x\rightarrow 0^+} \sqrt{x}=0\].

EXAMPLE 1

Use the Basic Limit Laws to evaluate:

(a) \(\lim\limits_{x\rightarrow 2} x^3\)
(b) \(\lim\limits_{x\rightarrow 2} (x^3 + 5x + 7)\)
(c) \(\lim\limits_{x\rightarrow 2} \sqrt{x^3 + 5x + 7}\)

Solution

(a) By Eq. (1), \(\lim\limits_{x\rightarrow 2} x^3 = 2^3 = 8\).
(b)\[\begin{aligned}\lim\limits_{x\rightarrow 2} (x^3 + 5x + 7) & = \lim\limits_{x\rightarrow 2} x^3 + \lim\limits_{x\rightarrow 2} 5x + \lim\limits_{x\rightarrow 2} 7 \text{ (Sum Law)} \\ &= \lim\limits_{x\rightarrow 2} x^3 + 5\lim\limits_{x\rightarrow 2} x + \lim\limits_{x\rightarrow 2} 7 \\ &=8+5(2) + 7 = 25\end{aligned}\]
(c) By Law (v) for roots and (b), \(\lim\limits_{x\rightarrow 2} \sqrt{x^3 + 5x + 7}=\sqrt{\lim\limits_{x\rightarrow 2} x^3 + 5x + 7} = \sqrt{25} = 5\)

Question 2.9 Limit Laws Progress Check 1

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Correct.

Try again. Apply the appropriate limit laws.

Incorrect.

EXAMPLE 2

Evaluate (a) \(\lim\limits_{t\rightarrow -1}\frac{t+6}{2t^4}\) and (b) \(\lim\limits_{t\rightarrow 3} t^{-\frac{1}{4}}(t+5)^{\frac{1}{3}}\).

Solution

(a) Use the Quotient, Sum, and Constant Multiple Laws:\[\lim\limits_{t\rightarrow -1}\dfrac{t+6}{2t^4}= \frac{\lim\limits_{t\rightarrow -1} (t+6)}{\lim\limits_{t\rightarrow -1}(2t^4)} = \frac{\lim\limits_{t\rightarrow -1} t+ \lim\limits_{t\rightarrow -1}6}{2\lim\limits_{t\rightarrow -1}t^4} = \frac{-1+6}{2(-1)^4} = \frac{5}{2}\]
You may have noticed that each of the limits in Examples 1 and 2 could have been evaluated by a simple substitution. For example, set \(t = -1\) to evaluate \(\lim\limits_{t\rightarrow -1}\frac{t+6}{2t^4} = \frac{-1+6}{2(-1)^4} = \frac{5}{2}\). Substitution is valid when the function is continuous, a concept we shall study in the next section.

(b) Use the Product, Powers, and Sum Laws: \begin{align*} \lim\limits_{t\rightarrow 3} t^{-\frac{1}{4}}(t+5)^{\frac{1}{3}} & =\left(\lim\limits_{t\rightarrow 3} t^{-\frac{1}{4}}\right)\left(\lim\limits_{t\rightarrow 3} \sqrt[3]{t+5}\right) = \left(3^{-\frac{1}{4}}\right)\left(\sqrt[3]{\lim\limits_{t\rightarrow 3} {t+5}}\right)\\[5pt] & = 3^{-\frac{1}{4}} \sqrt[3]{3+5} = 3^{-\frac{1}{4}}(2) = \frac{2}{3^{{1}/{4}}} \end{align*}

Question 2.10 Limit Laws Progress Check 2

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Correct.

Try again. Apply the appropriate limit laws.

Incorrect.