finite mathematics & applied calculus topic summary:
Introduction to the derivative: Part 1 of 3: Limits
Tools:

Limits

Let $f$ be a function such that $f(x)$ is defined for $x$ arbitrarily close to but different from $a.$ Then, if $f(x)$ gets arbitrarily close to the number $L$ for $x$ sufficiently close to (but not equal to) $a$ regardless from which side, we say that $f(x)$ approaches $\bold{L}$ as $\bold{x \to a}$ ("$x$ approaches $a$") or that the limit of $f(x)$ as $x \to a$ is $L$.

In other words, we can make $f(x)$ be as close to $L$ as we like by choosing any $x$ in the domain of $f$ sufficiently close to but not equal to $a.$ We write
$\displaystyle \lim_{x \to a} f(x) = L$
or
$f(x) \to L$ as $x \to a.$        The limit of $f(x)$ as $x \to a$ is $L$.
If $f(x)$ fails to approach a single fixed number as $x$ approaches $a$, then we say that $f(x)$ has no limit as $x \to a$, or
$\displaystyle \lim_{x \to a} f(x)\$ does not exist.
Examples: Limits

As $x$ approaches $2,$ $x+3$ approaches $2+3 = 5.$ That is, $x+3 \to 5$ as $x \to 2$, or
$\displaystyle \lim_{x \to 2} (x+3) = 5.$        The limit of $x+3$ as $x \to 2$ is $5$.

As $x$ approaches $0,$ $\sqrt{x}$ approaches $\sqrt{0}=0.$ That is, $\sqrt{x} \to 0$ as $x \to 0$, or
$\displaystyle \lim_{x \to 0}\sqrt{x} = 0.$        The limit of $\sqrt{x}$ as $x \to 0$ is $0$.
Note that, even though $\sqrt{x}$ is not defined to the left of $0$, it still satisfies the definition: $\sqrt{x}$ will be as close as you like to $0$ if $x$ is any number in the domain of the function sufficiently close to $0.$

Practice:

Estimating limits numerically

To numerically estimate $\displaystyle \lim_{x \to a} f(x)$ if it exists:
• Make a table of values for $f(x)$ using values of $x$ that approach $a$ closely from either side.
• If the limit exists, then the values of $f(x)$ will approach the limit as $x$ approaches $a$ from both sides.
• The more accurately you wish to evaluate this limit, the closer to $a$ you will need to choose the values of $x$.
Examples: Estimating limits numerically

To estimate $\displaystyle \lim_{x \to 3} \frac{x^2-9}{x-3},$ we make tables of values with $x$ approaching 3 from both sides:
$x$ approaching 3 from the left →
← $x$ approaching 3 from the right

Because the values of $f(x)$ appear to be approaching $6$ as $x$ approaches $3$ from either side, we estimate that the limit is $6.$

Practice:

One-sided limits

Let $f$ be a function such that $f(x)$ is defined for $x$ arbitrarily close to and on the left of $a.$ Then, if $f(x)$ gets arbitrarily close to the number $L$ for $x$ sufficiently close and on the left of $a$, we say that $f(x)$ approaches $\bold{L}$ as $\bold{x \to a^-}$ ("$x$ approaches $a$ from the left") or that the limit of $f(x)$ as $x \to a^-$ is $L$.

In other words, we can make $f(x)$ be as close to $L$ as we like by choosing any $x$ in the domain of $f$ sufficiently close to and to the left of $a.$ We write
$\displaystyle \lim_{x \to a^-} f(x) = L$
or
$f(x) \to L$ as $x \to a^-.$
The limit of $f(x)$ as $x \to a$ from the left is $L$.
Similarly, if $f(x)$ is defined for $x$ arbitrarily close to and on the right of $a$ and $f(x)$ gets arbitrarily close to $L$ for $x$ sufficiently close and on the right of $a$, we say that $f(x)$ approaches $\bold{L}$ as $\bold{x \to a^+}$ ("$x$ approaches $a$ from the right") or that the limit of $f(x)$ as $x \to a^+$ is $L$. In other words, we can make $f(x)$ be as close to $L$ as we like by choosing any $x$ in the domain of $f$ sufficiently close to and to the right of $a,$ and we write
$\displaystyle \lim_{x \to a^+} f(x) = L$
or
$f(x) \to L$ as $x \to a^+.$
The limit of $f(x)$ as $x \to a$ from the right is $L$.
If $f(x)$ approaches the same limit $L$ as $a \to a^-$ and as $x \to a^+$, then $\displaystyle \lim_{x \to a}f(x) = L$ as well:
$\displaystyle \lim_{x \to a}f(x) = \lim_{x \to a^-}f(x) = \lim_{x \to a^+}f(x) = L.$
If $f(x)$ approaches different limits as $a \to a^-$ and as $x \to a^+$, then $\displaystyle \lim_{x \to a}f(x)$ does not exist.
Examples: One-sided limits

$\displaystyle \lim_{x \to 2^-} (x+3) = 5. \qquad \lim_{x \to 2^+} (x+3) = 5.$

As $x$ approaches $0$ from the right, $\sqrt{x}$ approaches $\sqrt{0}=0.$ That is, $\sqrt{x} \to 0$ as $x \to 0^+$, or
$\displaystyle \lim_{x \to 0^+}\sqrt{x} = 0.$
Note that $\displaystyle \lim_{x \to 0^-}\sqrt{x}$ does not exist because $\sqrt{x}$ is not defined to the left of $0$. However, as we saw above, $\displaystyle \lim_{x \to 0}\sqrt{x}$ does exist. Thus
$\displaystyle \lim_{x \to 0}\sqrt{x} = \lim_{x \to 0^+}\sqrt{x} = 0,$ and

$\displaystyle \lim_{x \to 0^-}\sqrt{x}$ does not exist.

To estimate $\displaystyle \lim_{x \to 0^-} \frac{|x|}{x}$ and $\displaystyle \lim_{x \to 0^+} \frac{|x|}{x},$ we make tables of values with $x$ approaching 0 from both sides:
$x$ approaching 3 from the left →
so $\qquad \displaystyle \lim_{x \to 0^-}\frac{|x|}{x} = -1$.
← $x$ approaching 3 from the right
so $\qquad \displaystyle \lim_{x \to 0^+}\frac{|x|}{x} = 1$.

Because the left-and right limits are different,
$\displaystyle \lim_{x \to 0}\frac{|x|}{x} \ \$ does not exist.

Practice:

Limits at infinity

Let $f$ be a function such that $f(x)$ is defined for $x$ arbitrarily large and positive. Then, if $f(x)$ gets arbitrarily close to the number $L$ for $x$ sufficiently large and positive, we say that $f(x)$ approaches $\bold{L}$ as $\bold{x \to +\infty}$ or that the limit of $f(x)$ as $x \to +\infty$ is $L$.

In other words, we can make $f(x)$ be as close to $L$ as we like by choosing any $x$ in the domain of $f$ sufficiently large and positive. We write
$\displaystyle \lim_{x \to +\infty} f(x) = L$
or
$f(x) \to L$ as $x \to +\infty.$
The limit of $f(x)$ as $x \to +\infty$ is $L$.
Similarly, if $f(x)$ is defined for $x$ arbitrarily large numerically but negative, and $f(x)$ gets arbitrarily close to $L$ for $x$ sufficiently large numericaly and negative, we say that $f(x)$ approaches $\bold{L}$ as $\bold{x \to -\infty}$ or that the limit of $f(x)$ as $x \to -\infty$ is $L$. In other words, we can make $f(x)$ be as close to $L$ as we like by choosing any $x$ in the domain of $f$ sufficiently large numerically and negative, and we write
$\displaystyle \lim_{x \to -\infty} f(x) = L$
or
$f(x) \to L$ as $x \to -\infty.$
The limit of $f(x)$ as $x \to -\infty$ is $L$.
Examples: Limits at infinity

As $x$ approaches $+\infty, \$ $\frac{1}{3x+2}$ approaches $0.$ That is,
$\displaystyle \lim_{x \to +\infty} \frac{1}{3x+2} = 0.$        The limit of $\frac{1}{3x+2}$ as $x \to +\infty$ is $0$.
Further, as $x$ approaches $-\infty, \$ $\frac{1}{3x+2}$ approaches $0$ as well. That is,
$\displaystyle \lim_{x \to -\infty} \frac{1}{3x+2} = 0.$        The limit of $\frac{1}{3x+2}$ as $x \to -\infty$ is $0$.

To estimate $\displaystyle \lim_{x \to +\infty} \frac{x^2-x+1}{2x^2-3},$ we make tables of values with $x$ approaching $+\infty$:
$x$ approaching $+\infty$ →

Because the values of $f(x)$ appear to be approaching $\tfrac{1}{2}$ as $x$ approaches $+\infty$, we estimate that
$\displaystyle \lim_{x \to +\infty} \frac{x^2-x+1}{2x^2-3} = \frac{1}{2}.$

To estimate $\displaystyle \lim_{x \to -\infty} \frac{x^2-x+1}{2x^2-3},$ we make tables of values with $x$ approaching $-\infty$:
$x$ approaching $-\infty$ →

Because the values of $f(x)$ appear to be approaching $\tfrac{1}{2}$ as $x$ approaches $-\infty$, we estimate that
$\displaystyle \lim_{x \to -\infty} \frac{x^2-x+1}{2x^2-3} = \frac{1}{2}$     as well.

Practice:

Estimating limits graphically

To decide whether $\displaystyle \lim_{x \to a}f(x)$ exists and to estimate its value if it does:
1. Draw the graph of $f(x)$ by hand or with graphing technology.
2. Position your pencil point (or the trace cursor) on a point of the graph to the left of $x = a$.
3. Move the point along the graph toward $x = a$ from the left, and read the $y$-coordinate as you go. The value that the $y$-coordinate approaches (if any) is the limit $\displaystyle \lim_{x \to a^{-}}f(x)$.
4. Repeat Steps 2 and 3, this time starting from a point on the graph to the right of $x = a$ and approaching $x = a$ along the graph from the right. The value that the $y$-coordinate approaches (if any) is $\displaystyle \lim_{x \to a^+}f(x).$
5. If the left and right limits both exist and have the same value $L$, then $\displaystyle \lim_{x \to a}f(x) = L.$ If their values differ, the limit does not exist. The value $f(a)$ has no relevance whatsoever.
6. To evaluate $\displaystyle \lim_{x \to +\infty}f(x)$, move the pencil point toward the far right of the graph, and estimate the value the $y$-coordinate approaches (if any). For $\displaystyle \lim_{x \to -\infty}f(x)$, move the pencil point toward the far left.
Example: Estimating limits graphically

Refer to the following graph of the function $f:$

Computing limits as xa algebraically

To decide algebraically whether $\displaystyle \lim_{x \to a}f(x)$ exists and to calculate its value if it does:
1. Check to see whether $f$ is a closed form function. These are functions that can be specified by a single formula involving constants, powers of $x,$ radicals, exponentials and logarithms, combined using arithmetic operations and composition of functions. If $f$ is a closed-form function, then:
1. If $a$ is in the domain of $f,$ then $\displaystyle \lim_{x \to a}f(x)$ exists and equals $f(a).$
2. If $a$ is not in the domain of $f,$ but $f$ can be reduced by simplification to a closed-form function $g$ that does have $a$ in its domain, then $\displaystyle \lim_{x \to a}f(x)$ exists and equals $g(a).$.
3. If $a$ is not in the domain of $f,$ and you cannot simplify the function as in (b), then simplify as much as possible, and estimate the limit by further analysis (for instance, if substitution leads to a determinate form like $k/0$).
2. If $f$ is not closed-form, and $a$ is a point at which the definition of $f$ changes, compute the left limit and right limit seperately, and check whether they agree.
Examples: Computing limits as xa algebraically

Consider the limit $\displaystyle \lim_{x \to 1}\frac{x^2-9}{x-3}.$
Notice that the function $f(x)=\frac{x^2-9}{x-3}$ is closed form, and $a = 1$ is in its domain. Therefore, the limit is obained by substituting $x = 1$ (point (a) above):.
$\displaystyle \lim_{x \to 1}\frac{x^2-9}{x-3} = \frac{1-9}{1-3} = 4.$

Now consider $\displaystyle \lim_{x \to 3}\frac{x^2-9}{x-3}.$
Although the function $f(x)=\frac{x^2-9}{x-3}$ is closed form, $a = 3$ is not in its domain, and so we need to first simplify $f(x)$ to reduce it to a function that does have $3$ in its domain:.
 $\displaystyle \lim_{x \to 3}\frac{x^2-9}{x-3}$ $= \lim_{x \to 3}\frac{(x-3)(x+3)}{x-3}$ Does not have $x = 3$ in its domain; it is a singular point. $= \lim_{x \to 3}(x+3)$ Has $x = 3$ in its domain $= 3+3 = 6$ Evaluate

Practice:

Computing limits at infinity algebraically

To decide algebraically whether $\displaystyle \lim_{x \to \pm\infty}f(x)$ exists and to calculate its value if it does:
Check to see whether $f(x)$ is a ratio of polynomials (that is, a rational function). If it is, then you can ignore all but the highest powers of $x$ in the numerator and denominator. This simpler function will have the same limit as $f.$
Examples: Computing limits at infinity algebraically

Consider the limit $\displaystyle \lim_{x \to +\infty}\frac{x^3+x^2-9}{2x^3-x-3}.$
Here $f(x)$ is a ratio of polynomials, so we ignore everything except the highest power of $x$ in the numerator and denominator:.
 $\displaystyle \lim_{x \to +\infty}\frac{x^3+x^2-9}{2x^3-x-3}$ $\displaystyle = \lim_{x \to +\infty}\frac{x^3}{2x^3}$ Ignore all but highest powers top and bottom. $\displaystyle = \lim_{x \to +\infty}\frac{1}{2}$ Cancel the $x^3.$ $\displaystyle = \frac{1}{2}$

Now consider $\displaystyle \lim_{x \to +\infty}\frac{5x^3+x^2-9}{2x^4-x^3-3}.$
 $\displaystyle \lim_{x \to +\infty}\frac{5x^3+x^2-9}{2x^4-x^3-3}$ $\displaystyle = \lim_{x \to +\infty}\frac{5x^3}{2x^4}$ Ignore all but highest powers top and bottom. $\displaystyle = \lim_{x \to +\infty}\frac{5}{2x}$ Cancel $x^3.$ $= 0$ As $x$ gets large, $5/(2x)$ gets close to zero.

On the other hand,
 $\displaystyle \lim_{x \to -\infty}\frac{5x^5+x^2-9}{2x^4-x^3-3}$ $\displaystyle = \lim_{x \to -\infty}\frac{5x^5}{2x^4}$ Ignore all but highest powers top and bottom. $\displaystyle = \lim_{x \to -\infty}\frac{5x}{2}$ Cancel $x^4.$ $= -\infty$ As $x$ gets large and negative, so does $5x/2.$

Practice: