Evaluating Limits with Tables and Graphs

At the end of the last section, we got to this problem: lim⁡x→3x2−9x−3\lim_{x \to 3} \dfrac{x^2-9}{x-3} What we meant by this is that we want to know what this fraction is doing for values of xx near 3.3. In this section, we'll start to unpack that idea and see how to answer this question.

Our first idea is this: what if we just plug in 33 for x?x? After all, if we want to know what's happening near x=3,x=3, where better to look than at x=3?x=3?

32−93−3=00\dfrac{3^2-9}{3-3} = \dfrac{0}{0}

It was a good idea, but we get to this impossible answer of 0/0,0/0, so we need to try something else. In fact, as we'll see later, getting the answer 0/00/0 is actually pretty helpful, because it will give us an idea of what to do next (once we have a few more tools available to us).

Ways to Evaluate Limits

To get an answer to a question involving limits, we'll generally approach the problem in one of three ways:

  1. Graphing the function
  2. Building a table
  3. Using algebra to simplify the function

We'll often use more than one of these techniques on the same problem to compare their results.

Let's try all three of these methods on this example, and see what we can discover.

Graphing

This function, f(x)=x2−9x−3,f(x) = \dfrac{x^2-9}{x-3}, is probably not one that you can draw off the top of your head (unless your precalculus skills are particularly strong), so for now we'll have to rely on a graphing calculator. If we do so, we can get a graph that looks like this:

Graph of a straight line with a slope of 1 and passing through (0,3)

Does that strike you as odd? Does it seem strange that the graph of this rational function ended up just being a straight line?

In fact, if you look closer, you might get an idea of why the graph looks like this: this line has a slope of 11 and a yy-intercept of 3,3, meaning it has the equation y=x+3.y=x+3.

We set out to graph f(x)=x2−9x−3f(x) = \dfrac{x^2-9}{x-3} and we got the graph of f(x)=x+3.f(x) = x+3.

You can probably see why this is the case, but in case you can't, let's turn to another of our tools for evaluating limits: algebra.

Algebra

Let's try simplifying the function f(x)=x2−9x−3.f(x) = \dfrac{x^2-9}{x-3}.

We can factor the numerator, which will give us a common factor in both parts of the fraction that we can cancel. f(x)=x2−9x−3=(x+3)(x−3)x−3=x+3\begin{align*} f(x) &= \dfrac{x^2-9}{x-3}\ &= \dfrac{(x+3)\cancel{(x-3)}}{\cancel{x-3}}\ &= x+3 \end{align*}

Unsurprisingly, this matches the graph we saw above, so all is good, because everything is consistent.

look closer

Let's take another look at that example. Specifically, is it true that these two functions are equal? x2−9x−3=x+3\dfrac{x^2-9}{x-3} = x+3 Everything you've learned in your algebra studies has said that yes, we can cancel things like this and still have something equal to what we started with.

But now we're ready for a more careful answer: it's not exactly true that these two functions are equal.

Why not? Well, what does it mean for two functions to be equal?

Two functions are equal if every input will give the same output for both functions.

Is this true for these two functions? Almost, but there's a glaring exception: when x=3,x=3, they give very different answers: one gives the output of 66 and the other gives an error when you try to divide by 0.

This is the only issue, so we could accurately say x2−9x−3=x+3, as long as x≠3\dfrac{x^2-9}{x-3} = x+3 \textrm{, as long as } x \neq 3

What does this mean for the graph? You may recall this from precalculus, but when a zero of the denominator can be canceled, that indicates a hole in the graph at that point. So the actual graph of this function looks like this:

Graph of a line with slope 1 and passing through (0,3). There is a hole in the graph at (3,6)

Note that the graphing calculator didn't recognize the hole in the graph, because it just calculated the value of the function at a bunch of xx values, plotted the results, and connected them.

Table

Let's tackle this same example by building a table, checking the answer for values of xx that get very close to 3 and looking for a pattern. At this point, with the graph we just drew, you can probably guess what answer we'll arrive at, but let's look at how to build a table before discussing the answer.

Our goal is to pick values that are very close to 3, and to do so in a pattern that we can keep extending. For instance, we won't use 3.4, 3.3, 3.2, etc., because we quickly run out of options. If instead we use values like 3.1, 3.01, 3.001, notice that we could continue this pattern indefinitely (in practice, three or four iterations is usually enough to get an answer).

Also, notice carefully that we want to use values on both sides of 3; we refer to this as taking the limit from both sides. What this means is that we'll be filling in a table like this one:

xx x2−9x−3\dfrac{x^2-9}{x-3}
3.1
3.01
3.001
2.9
2.99
2.999

In practice, filling out this table is simple, but tedious (which makes it ideal for computers to do for us). All we have to do is plug each value of xx into the expression for the function and get the result.

If you take a minute to do so, you should get a table that looks like this:

xx x2−9x−3\dfrac{x^2-9}{x-3}
3.1 6.1
3.01 6.01
3.001 6.001
2.9 5.9
2.99 5.99
2.999 5.999

Here's the important bit: make sure that you can read through this table and see the trend. As the values of xx approach 3, the right-hand column shows that the values of f(x)f(x) get closer and closer to 6, and we imagine that this trend will continue.

Because of this, we can say lim⁡x→3x2−9x−3=6\lim_{x \to 3} \dfrac{x^2-9}{x-3} = 6 Notice how the notation works: under the word lim⁡\lim we indicate what we want xx to approach, then we write the function that we're evaluating, and after the equals sign we write our conclusion for what the function values (yy-values) approach as xx approaches 3.

We could read this expression this way: the limit as xx approaches 3 of (x2−9)/(x−3)(x^2-9)/(x-3) is 6. We could also write something like as x→3, f(x)→6.\textrm{as } x \to 3,\ f(x) \to 6.

Putting it all together

We'll focus more on the algebra in the next section, but for now, notice that we have three tools that we can use on this same example, and we can arrive at the answer using any of the three (although it's helpful to compare multiple methods, as we found).

In short, this function, (x2−9)/(x−3),(x^2-9)/(x-3), has a relatively simple graph: it looks like a straight line with a hole at (3,6).(3,6). When we simplified it algebraically, we reduced it to the equation of that line, and we noticed that canceling the term x−3x-3 indicated that there is a hole in the graph at x=3.x=3. Without the hole, the line would continue through the point (3,6).(3,6). Finally, the table showed us that as we input values of xx closer to 3,3,, the outputs inched closer to 6,6, making the results of all three methods consistent.

Comparing Methods

  1. Drawing a graph is often a great way to get an idea of what is happening as we approach different values with limits, but it's often difficult to draw a graph without a calculator. When it's convenient, though, we'll use a graph as much as we can.
  2. Building a table is simple, but pretty tedious, and it usually requires the use of a calculator. Because of this, we'll only use tables when we really have to; we'll use the other methods much more.
  3. Simplifying algebraically is the most common approach for evaluating limits, and we'll discuss this more in the next section.

One-Sided Limits

Remember when we were building the table to evaluate lim⁡x→3x2−9x−3\lim_{x \to 3} \dfrac{x^2-9}{x-3} we plugged in values close to 3 on both sides? What if we had only checked on one side?

There are times that we choose to do this, especially if the two sides lead to different answers. Let's take a look at an example of this.

Example 1One-Sided Limits

Use limits to describe what's happening near x=5x=5 for the function f(x)=4x−5f(x) = \dfrac{4}{x-5}

Graph

Let's start with the graph. Notice that this function is undefined for x=5,x=5, and the factor (x−5)(x-5) can't be canceled, which means that there is a vertical asymptote in the graph at x=5.x=5.

Also, notice that the base function for this one is 1/x,1/x, which has this graph: The graph of the reciprocal function 1/x

Changing xx to x−5x-5 shifts the graph to the right by 5,5, and multiplying by 44 stretches it vertically, so our graph looks like this: The graph of 4/(x-5), which has a vertical asymptote at 5.

If we plug in values of xx close to 55 for this function, we'll get different answers for inputs below 55 and those above 5.5. Notice that plugging in values lower than 55 will give us negative answers that are going downward rapidly (heading toward −∞-\infty) and values higher than 55 will give answers heading toward +∞.+\infty.

Because of this, we'll identify the one-sided limits separately: lim⁡x→5−4x−5=−∞lim⁡x→5+4x−5=∞\begin{align*} \lim_{x \to 5^-} \dfrac{4}{x-5} &= -\infty\ \lim_{x \to 5^+} \dfrac{4}{x-5} &= \infty \end{align*} Notice the small superscripts on the 55 in the limit notation: the minus sign indicates that we're approaching 55 from the left (values less than 55), and the plus sign means we're approaching 55 from the right.

Table

Let's see how we can observe the same thing from a table (if, for instance, we struggled to draw the graph by hand).

If we set up our table the same way we did before, we'd start with xx values like 5.1, 5.01, etc. from the right and values like 4.9, 4.99, etc. from the left.

The resulting table would look like this:

xx 4x−5\dfrac{4}{x-5}
5.1 40
5.01 400
5.001 4000
4.9 -40
4.99 -400
4.999 -4000

Make sure you can read the trend in the right-hand column. Notice that as we approach 55 from the right, the results go from 40 to 400 to 4000, which indicates unlimited growth, trending toward ∞,\infty, so lim⁡x→5+4x−5=∞\lim_{x \to 5^+} \dfrac{4}{x-5} = \infty

Similarly, as we plug in values closer to 55 from the left, the output grows in the negative direction, from -40 to -400 to -4000, trending toward −∞,-\infty, so lim⁡x→5−4x−5=−∞\lim_{x \to 5^-} \dfrac{4}{x-5} = -\infty

What do we do when the right- and left-sided limits don't agree?

Limit Does Not Exist (DNE)

If the one-sided limits do not give the same answer, the two-sided limit (without the plus/minus in the limit notation) does not exist (often abbreviated DNE).

If lim⁡x→c+f(x)≠lim⁡x→c−f(x)\displaystyle\lim_{x \to c^+} f(x) \neq \displaystyle\lim_{x \to c^-} f(x), then lim⁡x→cf(x)\displaystyle\lim_{x \to c} f(x) DNE.

Because of this, we generally only talk about the one-sided limits when they differ. If they're equal, we'll just refer to the overall limit (which we can call the two-sided limit), and recognize in that case that the answer for the overall limit is the same as the answer for both one-sided limits.

Example 2Comparing Limit Methods

Consider the function f(x)=x+1x2+3x+2f(x) = \dfrac{x+1}{x^2+3x+2} Use all three methods described in this section to describe the function, and specifically use limits to investigate how the function behaves near

  1. x=−1x=-1
  2. x=−2x=-2
  3. x=1x=1

as well as what happens as xx approaches ∞\infty and −∞.-\infty.

Algebra

Let's begin by simplifying the function, because if we pay attention as we do so, we can learn a lot about its graph. x+1x2+3x+2=x+1(x+1)(x+2)=x+1(x+1)(x+2)=1x+2, as long as x≠−1\begin{align*} \dfrac{x+1}{x^2+3x+2} &= \dfrac{x+1}{(x+1)(x+2)}\ &= \dfrac{\cancel{x+1}}{\cancel{(x+1)}(x+2)}\ &= \dfrac{1}{x+2} \textrm{, as long as } x \neq -1 \end{align*}

By factoring the denominator, we found two zeros: at x=−1x=-1 and x=−2x=-2. The one at x=−1x=-1 was removed by canceling a matching factor from the numerator, which tells us that there is a hole in the graph at x=−1.x=-1. Because x+2x+2 was not canceled away, we know that there is a vertical asymptote at x=−2.x=-2.

Furthermore, the simplified version is 1/(x+2),1/(x+2), which is the result of shifting 1/x1/x to the left by 2.2.

Graph

Based on these observations, we can draw the graph of f(x)f(x) with a fair amount of detail. The graph below was done with a calculator, but we could have drawn a rough sketch of this with what we gleaned above. Graph of the function in this example, with a vertical asymptote at -2 and a hole at -1

Table

We could build separate tables for each of the limits requested, but for the sake of space (and sanity), we'll just consider two here: one as xx approaches −1,-1, and another as xx approaches ∞.\infty. For the second table, we'll just need to pick values of xx that are increasing rapidly; for example: 10, 100, 1000, etc.

Here's the first table:

xx x+1x2+3x+2\dfrac{x+1}{x^2+3x+2}
-0.9 0.9091
-0.99 0.9901
-0.999 0.9990
-1.1 1.1111
-1.01 1.0101
-1.001 1.0010

And the second:

xx x+1x2+3x+2\dfrac{x+1}{x^2+3x+2}
10 0.0833
100 0.0098
1000 0.0010

Note that with xx approaching ∞,\infty, there's no sense of approaching from both sides, because there's nothing to the right of ∞.\infty.

We'll use these tables as we answer the questions posed about this function.

Evaluating Limits

Let's use what we found above to evaluate the limits we were given.

  1. As xx approaches −1-1: based on the graph, we can observe a hole in the graph at (−1,1),(-1,1), so as xx approaches −1-1 from either side, the values of f(x)f(x) will approach 1.1. This is confirmed by the table we built; the values in the right-hand column got closer to 11 as xx got closer to −1-1 from each side. lim⁡x→−1x+1x2+3x+2=1\boxed{\lim_{x \to -1} \dfrac{x+1}{x^2+3x+2} = 1} Note that we get the same result if we plug x=−1x = -1 into the simplified form of f(x),f(x), because simplifying the function by canceling x+1x+1 has the same result as filling in the hole in the graph.

  2. To see what is happening around x=−2,x=-2, it's best to look at the graph. We see a vertical asymptote there, and we notice that on the right of the asymptote the graph is curving upward toward ∞\infty and on the left, it's going downward toward −∞.-\infty. Therefore we conclude the following: lim⁡x→−2+f(x)=∞lim⁡x→−2−f(x)=−∞lim⁡x→−2f(x)  DNE\boxed{\begin{align*} \lim_{x \to -2^+} f(x) &= \infty\ \lim_{x \to -2^-} f(x) &= -\infty\ \lim_{x \to -2} f(x) &\ \textrm{ DNE}\ \end{align*}}

  3. Take a look at the graph near x=1:x=1: there's nothing unusual happening here. The graph simply passes through that section without any breaks, holes, asymptotes, or anything else of interest. Because of that, we don't really need to do anything fancy with the limit (although we could build a table if we really wanted to waste some time). We can simply evaluate the function at x=1x=1 and call it a day: lim⁡x→1f(x)=f(1)=13\boxed{\lim_{x \to 1} f(x) = f(1) = \dfrac{1}{3}}

  4. Finally, let's consider what happens as xx approaches positive or negative infinity. On the graph, this means looking at the end behavior, out to the right and left edges of the graph: Graph of the function in this example, with a vertical asymptote at -2 and a hole at -1 At both ends of the graph, the function values (yy-values) are approaching the horizontal asymptote of 0 (one from above and one from below). Thus, lim⁡x→∞f(x)=lim⁡x→−∞f(x)=0\boxed{\lim_{x \to \infty} f(x) = \lim_{x \to -\infty} f(x) = 0} Note that this is consistent with what we observed in the table for xx approaching ∞.\infty.

That was a long example, but it illustrates all of the tools we have at our disposal to think about the behavior of limits. Generally we'll only use one method at a time, though, so we won't have so much redundancy in one problem.

Let's wrap up this section with a few short examples to illustrate some special cases.

Example 3Limit of a Polynomial

Evaluate the following limit: lim⁡x→3x2−2x+4\lim_{x \to 3} x^2-2x+4

Solution

We can approach this one the same way as the previous ones, but let's try something a little different: let's try imagining the result we'd get by looking at a graph or table without actually drawing either of them.

What do we know about the graph of a polynomial? You may have forgotten this from precalculus, but every polynomial has a smooth, continuous graph; the most interesting thing about the graph tends to be its turning points and intercepts, but those aren't relevant here.

The fact that the graph is continuous, without any holes or asymptotes, means that it behaves like the limit as x→1x \to 1 in the last example. Therefore, we can evaluate this limit simply by plugging 33 in for x:x: lim⁡x→3x2−2x+4=(3)2−2(3)+4=7\begin{align*} \lim_{x \to 3} x^2-2x+4 &= (3)^2 - 2(3) + 4\ &= \boxed{7} \end{align*}

We'll revisit this idea in later sections, but for now, we'll simply draw the conclusion that for polynomials, we can evaluate limits by substitution; no fancy work is needed.

Example 4Limit of a Square Root Function

Evaluate the following limit: lim⁡x→2x−2\lim_{x \to 2} \sqrt{x-2}

Solution

You may be able to reason this one out, but more likely, you'll need to consider the graph.

In case you've forgotten, the graph of the square root function looks like half of a parabola on its side (because it's the inverse of x2x^2), and this one has been shifted to the right by 2 (because of the x−2x-2 inside the root). The graph of the square root of (x-2)

Now consider the limit as xx approaches 2.2. This one may seem obvious; it looks like the function approaches 0,0, right?

However, think carefully about how we evaluate limits: remember that we need the limits on both sides to agree in order to have an overall (two-sided) answer for the limit.

What's happening as xx approaches 22 from the left? Since the function isn't defined there, the answer is that the left-sided limit doesn't exist. Therefore, the right- and left-sided limits don't agree, meaning the two-sided limit doesn't exist.

We could summarize the answer this way: lim⁡x→2+x−2=0lim⁡x→2−x−2  DNElim⁡x→2x−2  DNE\begin{align*} \lim_{x \to 2^+} \sqrt{x-2} &= 0\ \lim_{x \to 2^-} \sqrt{x-2} &\ \textrm{ DNE}\ \lim_{x \to 2} \sqrt{x-2} &\ \textrm{ DNE}\ \end{align*}

Example 5Limit of a Piecewise Function

Consider the following function: f(x)={−2x+4 if x≤−18 if −1<x≤3x2−1 if 3<xf(x) = \left{\begin{array}{c l} -2x+4 & \textrm{ if } x \leq -1\ 8 & \textrm{ if } -1 < x \leq 3\ x^2-1 & \textrm{ if } 3 < x \end{array}\right. Find the limits as xx approaches −1-1 and 3.3.

Graph

Let's start with the graph. To graph a piecewise-defined function, like this one, imagine drawing each definition separately, but only including the section of it that's in the right domain.

For instance, the first like, −2x+4,-2x+4, describes a line with a slope of −2-2 and a yy-intercept of 4.4. We won't draw this whole line, but simply the portion of it to the left of x=−1.x=-1.

If it helps, we could figure out that the yy-value at x=−1x=-1 would be 6,6, so that section starts at the point (−1,6)(-1,6) and continues to the left, going up 22 steps for every step to the left.

After doing this with all three sections, we get the following graph: The graph of the piecewise-defined function described in this example.

Notice at x=−1,x=-1, the two sides of the graph don't come to the same point, meaning that the left- and right-sided limits don't agree: lim⁡x→−1−f(x)=6lim⁡x→−1+f(x)=8lim⁡x→−1f(x)  DNE\begin{align*} \lim_{x \to -1^-} f(x) &= 6\ \lim_{x \to -1^+} f(x) &= 8\ \lim_{x \to -1} f(x) &\ \textrm{ DNE}\ \end{align*}

At x=3,x=3, on the other hand, the two branches of the graph meet at a single point, so the limit does exist there: lim⁡x→3f(x)=8\lim_{x \to 3} f(x) = 8

Table

We're not actually going to build a table here, but simply imagine what would happen if we did.

Let's start around x=−1.x = -1. From the left, we would pick values slightly smaller than −1-1 and plug them into the first line: −2x+4-2x+4 Unsurprisingly, the answers we would get from doing so would match what we could get by simply plugging −1-1 into this expression: −2(−1)+4=6-2(-1) + 4 = 6 Thus we conclude that lim⁡x→−1−f(x)=6\lim_{x \to -1^-} f(x) = 6 (if you don't believe me, try some specific values)

To get the limit from the right, we can imagine evaluating the function (using the second line, where f(x)=8f(x) = 8) for values of xx slightly larger than −1,-1, which would all give us 88 as the result. We would then conclude that lim⁡x→−1+f(x)=8\lim_{x \to -1^+} f(x) = 8

Now try doing this for values to the left and right of x=3x=3 and see if you can replicate the answers we found from the graph.

For a piecewise-defined function, then, we can draw a conclusion from this example. As long as the individual sections of the function don't have holes or asymptotes, taking the limit at one of the breakpoints in the function is as simple as substituting xx into the definition from the left and right and seeing if the answers agree.

Example 6An Odd Example

Consider the following limit: lim⁡x→0sin⁡(1x)\lim_{x \to 0} \sin\left(\dfrac{1}{x}\right) (note: in this course, unless otherwise specified, we'll assume that all trig functions use radians)

We could try building a table for this limit, but it won't be very helpful.

xx sin⁡(1x)\sin\left(\dfrac{1}{x}\right)
0.1 -0.54
0.01 -0.51
0.001 0.83
-0.1 0.54
-0.01 0.51
-0.001 -0.83

Do you see any pattern there? I sure don't.

Okay, let's take a look at the graph instead. Of course, the challenge here is knowing how to graph this function. It's probably not one that you've seen before, so we really need to rely on a graphing calculator. The graph looks like this: The graph of the sine of 1 over x. The graph oscillates quickly near 0.

Notice how the function starts oscillating wildly near x=0.x=0. If you zoom in, you can see that this oscillation speeds up the closer you get to the middle: A zoomed in graph of the sine of 1 over x, showing the speed of oscillation increasing near 0.

Because of this, building a table is hopeless, because we'd just find values randomly placed along one of these oscillating curves.

What is the answer for the limit, then? Well, notice that this function bounces back and forth between 11 and −1,-1, moving faster and faster, but never settling down to a single value.

Since there's no single value that it approaches, we would say that this limit does not exist.

As we found in that last example, building tables and drawing graphs will always have their limitations. They are great tools to have available, but we can't rely on them every time.

Besides the fact that building these tables or graphs is tedious, they need the answer to be a nice round number. For instance, if the answer to a limit was e2−1,e^2-1, is there any chance we'd be able to recognize that from the values in a table, or spot that point on a graph?

For all these reasons, we'll turn our attention in the next section to some algebraic techniques that we can reliably use to evaluate limits in general.

Homework: Evaluating Limits with Tables and Graphs

  1. Use this graph of f(x)f(x) to calculate the limits below. Graph of x^2 with a hole at (2,4) and a solid dot at (2,1)

    1. lim⁡x→0f(x)\displaystyle\lim_{x \to 0} f(x)
    2. lim⁡x→−1f(x)\displaystyle\lim_{x \to -1} f(x)
    3. lim⁡x→2f(x)\displaystyle\lim_{x \to 2} f(x)
    4. lim⁡x→3−f(x)\displaystyle\lim_{x \to 3^-} f(x)
    5. lim⁡x→−2+f(x)\displaystyle\lim_{x \to -2^+} f(x)

  2. Use this graph of g(x)g(x) to calculate the limits below. Graph of a piecewise function defined by x+6 before x is -1, defined by x^3-2x^2-x+7 between -1 and 2, and -x^2+5x-3 after 2. There is a solid dot at (2,3) and an empty circle at (2,5)

    1. lim⁡x→−1−g(x)\displaystyle\lim_{x \to -1^-} g(x)
    2. lim⁡x→−1+g(x)\displaystyle\lim_{x \to -1^+} g(x)
    3. lim⁡x→−1g(x)\displaystyle\lim_{x \to -1} g(x)
    4. lim⁡x→2−g(x)\displaystyle\lim_{x \to 2^-} g(x)
    5. lim⁡x→2+g(x)\displaystyle\lim_{x \to 2^+} g(x)
    6. lim⁡x→2g(x)\displaystyle\lim_{x \to 2} g(x)
    7. lim⁡x→−3−g(x)\displaystyle\lim_{x \to -3^-} g(x)
    8. lim⁡x→−3+g(x)\displaystyle\lim_{x \to -3^+} g(x)
    9. lim⁡x→−3g(x)\displaystyle\lim_{x \to -3} g(x)

  3. Build a table (using a calculator) to estimate the value of each of the following limits.

    1. lim⁡x→2f(x)\displaystyle\lim_{x \to 2} f(x) if f(x)=5x−7f(x) = 5x-7
    2. lim⁡x→3x2−9x+3\displaystyle\lim_{x \to 3} \dfrac{x^2-9}{x+3}
    3. lim⁡x→0sin⁡(3x)2x\displaystyle\lim_{x \to 0} \dfrac{\sin(3x)}{2x}
    4. lim⁡x→1∣x2−1∣x−1\displaystyle\lim_{x \to 1} \dfrac{|x^2-1|}{x-1}
    5. lim⁡x→4x−2x−4\displaystyle\lim_{x \to 4} \dfrac{\sqrt{x} - 2}{x-4}

  4. For the following function, draw a graph and use that to evaluate lim⁡x→−4−f(x),\displaystyle\lim_{x \to -4^-} f(x), lim⁡x→−4+f(x),\displaystyle\lim_{x \to -4^+} f(x), and lim⁡x→−4f(x).\displaystyle\lim_{x \to -4} f(x). f(x)={−2x−2 if x<−4x if x≥−4f(x) = \left{\begin{array}{r l} -2x-2 & \textrm{ if } x < -4\ x & \textrm{ if } x \geq -4 \end{array}\right.

  5. For the following function, build a table and use that to evaluate lim⁡x→−5−f(x),\displaystyle\lim_{x \to -5^-} f(x), lim⁡x→−5+f(x),\displaystyle\lim_{x \to -5^+} f(x), and lim⁡x→−5f(x).\displaystyle\lim_{x \to -5} f(x). f(x)={3x2 if x<−54x if x≥−5f(x) = \left{\begin{array}{r l} 3x^2 & \textrm{ if } x < -5\ 4x & \textrm{ if } x \geq -5 \end{array}\right.

  6. Use a table or a graph to evaluate each of the following limits:

    1. lim⁡x→3x2−3xx2−2x−3\displaystyle\lim_{x \to 3} \dfrac{x^2-3x}{x^2-2x-3}
    2. lim⁡x→0∣x∣x\displaystyle\lim_{x \to 0} \dfrac{|x|}{x}
    3. lim⁡x→2x−2\displaystyle\lim_{x \to 2} \sqrt{x-2}

Homework Answers

  1. (a) 0 (b) 1 (c) 4 (d) 9 (e) 4
  2. (a) 5 (b) 5 (c) 5 (d) 5 (e) 3 (f) DNE (g) 3 (h) 3 (i) 3
  3. (a) 3 (b) 0 (c) 32\dfrac{3}{2} (d) DNE (e) 14\dfrac{1}{4}
  4. lim⁡x→−4−f(x)=6,\displaystyle\lim_{x \to -4^-} f(x) = 6, lim⁡x→−4+f(x)=−4,\displaystyle\lim_{x \to -4^+} f(x) = -4, and lim⁡x→−4f(x)\displaystyle\lim_{x \to -4} f(x) DNE
  5. lim⁡x→−5−f(x)=75,\displaystyle\lim_{x \to -5^-} f(x) = 75, lim⁡x→−5+f(x)=−20,\displaystyle\lim_{x \to -5^+} f(x) = -20, and lim⁡x→−5f(x)\displaystyle\lim_{x \to -5} f(x) DNE
  6. (a) 34\dfrac{3}{4} (b) DNE (c) DNE