MA005: Calculus I

Calculus has been called the study of continuous change, and the limit is the basic concept which allows us to describe and analyze such change. An understanding of limits is necessary to understand derivatives, integrals, and other fundamental topics of calculus.

The limit of a function describes the behavior of the function when the variable is near, but does not equal, a specified number (Fig. 1). If the values of $\mathrm{f}(\mathrm{x})$ get closer and closer, as close as we want, to one number $L$ as we take values of $x$ very close to (but not equal to) a number $\mathrm{c}$, then we 

$f(c)$ is a single number that describes the behavior (value) of f AT the point $x=c$.

$\lim\limits_{x \rightarrow c} \mathbf{f}(\mathbf{x}) \quad$ is a single number that describes the behavior of $f$ NEAR, BUT NOT AT, the point $\mathrm{x}=\mathrm{c}$.

If we have a graph of the function near $x=c$, then it is usually easy to determine $\lim\limits_{x \rightarrow c} f(x)$.

Example 1: Use the graph of $\mathrm{y}=\mathrm{f}(\mathrm{x})$ in Fig. 2 to determine the following limits:

(a) $\lim\limits_{x \rightarrow 1} \mathrm{f}(\mathrm{x})$

(b) $\lim\limits_{x \rightarrow 2} \mathrm{f}(\mathrm{x})$

(c) $\lim\limits_{x \rightarrow 3} \mathrm{f}(\mathrm{x})$

(d) $\lim\limits_{x \rightarrow 4} \mathrm{f}(\mathrm{x})$

Solution: (a) $\quad \lim\limits_{x \rightarrow 1} \mathrm{f}(\mathrm{x})=2$. When $\mathrm{x}$ is very close to 1 , the values of $\mathrm{f}(\mathrm{x})$ are very close to $\mathrm{y}=2 .$ In this example, it happens that $\mathrm{f}(1)=2$, but that is irrelevant for the limit. The only thing that matters is what happens for $x$ close to 1 but $x \neq 1$.

(b) $\mathrm{f}(2)$ is undefined, but we only care about the behavior of $\mathrm{f}(\mathrm{x})$ for $\mathrm{x}$ close to 2 and not equal to 2. When $\mathrm{x}$ is close to 2 , the values of $\mathrm{f}(\mathrm{x})$ are close to 3 . If we restrict $\mathrm{x}$ close enough to 2 , the values of $y$ will be as close to 3 as we want, so $\lim\limits_{x \rightarrow 2} f(x)=3$.

(c) When $\mathrm{x}$ is close to 3 (or as $\mathrm{x}$ approaches the value 3 ), the values of $\mathrm{f}(\mathrm{x})$ are close to 1 (or approach the value 1), so $\lim\limits_{x \rightarrow 3} \mathrm{f}(\mathrm{x})=1$. For this limit it is completely irrelevant that $\mathrm{f}(3)=2$, We only care about what happens to $\mathrm{f}(\mathrm{x})$ for $\mathrm{x}$ close to and not equal to 3.

(d) This one is harder and we need to be careful. When $x$ is close to 4 and slightly less than 4 ($x$ is just to the left of 4 on the $x$-axis), then the values of $f(x)$ are close to $2.$ But if $x$ is close to 4 and slightly larger than 4 then the values of $f(x)$ are close to $3.$ If we only know that $x$ is very close to 4, then we cannot say whether $y=f(x)$ will be close to 2 or close to 3 - it depends on whether $x$ is on the right or the left side of $4.$ In this situation, the $f(x)$ values are not close to a single number so we say $\lim\limits_{x \rightarrow 4} \mathrm{f}(\mathrm{x})$ does not exist. It is irrelevant that $\mathrm{f}(4)=1.$ The limit, as $\mathrm{x}$ approaches 4, would still be undefined if $\mathrm{f}(4)$ was 3 or 2 or anything else.

Practice 1: Use the graph of $\mathrm{y}=\mathrm{f}(\mathrm{x})$ in Fig. 3 to determine the following limits:

(a) $\lim\limits_{x \rightarrow 1} \mathrm{f}(\mathrm{x})$

(b) $\lim\limits_{t \rightarrow 2} \mathrm{f}(\mathrm{t})$

(c) $\lim\limits_{x \rightarrow 3} \mathrm{f}(\mathrm{x})$

(d) $\lim\limits_{w \rightarrow 4} \mathrm{f}(\mathrm{w})$

Example 2: Determine the value of $\lim\limits_{x \rightarrow 3} \frac{2 x^{2}-x-1}{x-1}$.

Solution: We need to investigate the values of $f(x)$ when $x$ is close to 3. If the $f(x)$ values get arbitrarily close to or even equal some number $\mathrm{L}$, then $\mathrm{L}$ will be the limit. One way to keep track of both the $x$ and the $f(x)$ values is to set up a table and to pick several $x$ values which are closer and closer (but not equal) to $3.$ We can pick some values of $x$ which approach 3 from the left, say $x=2.91,2.9997$ $2.999993$, and $2.9999999$, and some values of $x$ which approach 3 from the right, say $x=3.1,3.004$ $3.0001$, and $3.000002$. The only thing important about these particular values for $x$ is that they get closer and closer to 3 without equaling $3.$ You should try some other values "close to 3" to see what happens.

Instead of using a table of values, we could have graphed $y=f(x)$ for $x$ close to 3, Fig. 4, and used the graph to answer the limit question. This graphic approach is easier, particularly if you have a calculator or computer do the graphing work for you, but it is really very similar to the "table of values" method: in each case you need to evaluate $\mathrm{y}=\mathrm{f}(\mathrm{x})$ at many values of $\mathrm{x}$ near 3.

You might have noticed that if we just evaluate $f(3)$, then we get the correct answer $7.$ That works for this particular problem, but it often fails. The next example illustrates the difficulty.

Example 3: Find $\lim\limits_{x \rightarrow 1} \frac{2 x^{2}-x-1}{x-1}$. (Same as Example 2 but with $\mathrm{x} \rightarrow 1$).

Solution: You might try to evaluate $f(x)=\frac{2 x^{2}-x-1}{x-1}$ at $x=1$, but $\mathrm{f}$ is not defined at $\mathrm{x}=1$. It is tempting, but wrong, to conclude that this function does not have a limit as $\mathrm{x}$ approaches 1.

Table Method: Trying some "test" values for $\mathrm{x}$ which get closer and closer to 1 from both the left and the right, we get

Graph Method: We can graph $y=f(x)=\frac{2 x^{2}-x-1}{x-1}$ for $x$ close to 1 , Fig. 5, and notice that whenever $\mathrm{x}$ is close to 1 , the values of $\mathrm{y}=\mathrm{f}(\mathrm{x})$ are close to $3 . \mathrm{f}$ is not defined at $\mathrm{x}=1$, so the graph has a hole above $x=1$, but we only care about what $f(x)$ is doing for $x$ close to but not equal to 1.

Algebra Method: We could have found the same result by noting that $f(x)=\frac{2 x^{2}-x-1}{x-1}=\frac{(2 x+1)(x-1)}{x-1}=2 x+1$ as long as $\mathrm{x} \neq 1$. (If $\mathrm{x} \neq 1$, then $\mathrm{x}-1 \neq 0$ so it is valid to divide the numerator and denominator by the factor $x-1.$) The "$x \rightarrow 1$" part of the limit means that $x$ is close to 1 but not equal to $1$, so our division step is valid and

$\lim\limits_{x \rightarrow 1} \frac{2 x^{2}-x-1}{x-1}=\lim _{x \rightarrow 1} 2 x+1=3$, the correct answer.

Source: Dave Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2011/11/2-2FunctionLimit.pdf
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