Properties of Limits

Read this section to learn about the properties of limits. Work through practice problems 1-6.


This section presents results which make it easier to calculate limits of combinations of functions or to show that a limit does not exist. The main result says we can determine the limit of "elementary combinations" of functions by calculating the limit of each function separately and recombining these results for our final answer. 

Main Limit Theorem:

If \quad \lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x})=\mathrm{L} and \lim\limits_{x \rightarrow a} \mathrm{~g}(\mathrm{x})=\mathrm{M},

then \text { (a) } \lim\limits_{x \rightarrow a}\{\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})\}=\lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x})+\lim\limits_{x \rightarrow a} \mathrm{~g}(\mathrm{x})=\mathrm{L}+\mathrm{M}

(b) \lim\limits_{x \rightarrow a}\{\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})\}=\lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x})-\lim\limits_{x \rightarrow a} \mathrm{~g}(\mathrm{x})=\mathrm{L}-\mathrm{M}

(c) \lim\limits_{x \rightarrow a} \mathrm{k} \,  \mathrm{f(x)} \quad=\mathrm{k} \lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x}) \quad=\mathrm{kL}

(d) \lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})=\left\{\lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x})\right\} \cdot\left\{\lim\limits_{x \rightarrow a} \mathrm{~g}(\mathrm{x})\right\}=\mathrm{L} \cdot

(e) \lim\limits_{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim\limits_{x \rightarrow a} f(x)}{\lim\limits_{x \rightarrow a} g(x)} \qquad \qquad \qquad =\frac{\mathrm{L}}{\mathrm{M}} \quad( if \mathrm{M} \neq 0)

(f) \lim\limits_{x \rightarrow a}\{\mathrm{f}(\mathrm{x})\}^{\mathrm{n}}\qquad =\left\{\lim\limits_{x \rightarrow a} \mathrm{f}(\mathrm{x})\right\}^{\mathrm{n}} \qquad \qquad =\mathrm{L}^{\mathrm{n}}

(g) \lim\limits_{x \rightarrow a} \sqrt[n]{f(x)}=\sqrt[n]{\lim\limits_{x \rightarrow a} f(x)} \qquad \qquad =\sqrt[n]{L} \quad (if L>0 when n is even)

The Main Limit Theorem says we get the same result if we first perform the algebra and then take the limit or if we take the limits first and then perform the algebra: e.g., (a) the limit of the sum equals the sum of the limits. A proof of the Main Limit Theorem is not inherently difficult, but it requires a more precise definition of the limit concept than we have given, and it then involves a number of technical difficulties.

Practice 1: For \mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-\mathrm{x}-6 and \mathrm{g}(\mathrm{x})=\mathrm{x}^{2}-2 \mathrm{x}-3, evaluate the following limits:

(a) \lim\limits_{x \rightarrow 1}\{\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})\}     (b) \lim\limits_{x \rightarrow 1} \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})     (c) \lim\limits_{x \rightarrow 1}
            \mathrm{f}(\mathrm{x}) / \mathrm{g}(\mathrm{x})     (d) \lim\limits_{x \rightarrow 3}\{\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})\}

(e) \lim\limits_{x \rightarrow 3} \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})              (f) \lim\limits_{x \rightarrow 3} \mathrm{f}(\mathrm{x}) / \mathrm{g}(\mathrm{x})    (g) \lim\limits_{x
            \rightarrow 2}\{\mathrm{f}(\mathrm{x})\}^{3}         (h) \lim\limits_{x \rightarrow 2} \sqrt{1-g(x)}

Source: Dave Hoffman,
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