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Discover Calculus Single-Variable Calculus Topics with Motivating Activities

Peter Keep

Section 1.4 Limits Involving Infinity

Two types of limits involving infinity. In both cases, we’ll mostly just consider what happens when we divide by small things and what happens when we divide by big things. We can summarize this here, though:
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Fractions with small denominators are big, and fractions with big denominators are small.
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Subsection Infinite Limits

When we talked about indeterminate forms in Section 1.3, we noticed that the function value wasn’t defined, since we divided by 0. Specifically, we were looking at the \(\frac{0}{0}\) form in the limit. What happens in other cases when we divide by 0 with a nonzero numerator?
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Activity 1.4.1. What Happens When We Divide by 0?

First, let’s make sure we’re clear on one thing: there is no real number than is represented as some other number divided by 0.
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When we talk about “dividing by 0” here (and in Section 1.3), we’re talking about the behavior of some function in a limit. We want to consider what it might look like to have a function that involves division where the denominator gets arbitrarily close to 0 (or, the limit of the denominator is 0).
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(a)
Remember when, once upon a time, you learned that dividing one a number by a fraction is the same as multiplying the first number by the reciprocal of the fraction? Why is this true?
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(b)
What is the relationship between a number and its reciprocal? How does the size of a number impact the size of the reciprocal? Why?
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(d)
Let’s again consider \(12\div N\text{.}\) What is the value of this division problem when:
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(e)
Consider a function \(f(x) = \frac{12}{x}\text{.}\) What happens to the value of this function when \(x\to 0^+\text{?}\) Note that this means that the \(x\)-values we’re considering most are very small and positive.
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(f)
Consider a function \(f(x) = \frac{12}{x}\text{.}\) What happens to the value of this function when \(x\to 0^-\text{?}\) Note that this means that the \(x\)-values we’re considering most are very small and negative.
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So we want to formalize this kind of behavior. We know that the limits that we’re looking at don’t exist (since there isn’t a single, real number that the function outputs are all close to), but there is definitely some sort of consistent behavior that we’d like to signify.
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Definition 1.4.1. Infinite Limit.

We say that a function \(f(x)\) has an infinite limit at \(a\) if \(f(x)\) is arbitrarily large (positive or negative) when \(x\) is sufficiently close to, but not equal to, \(x=a\text{.}\)
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We would then say, depending on the sign of the values of \(f(x)\text{,}\) that:
\begin{equation*} \lim_{x\to a^-} f(x) = \pm \infty \hspace{3cm} \lim_{x\to a^+} f(x) = \pm\infty\text{.} \end{equation*}
If the sign of both one-sided limits are the same, we can say that \(\displaystyle \lim_{x\to a} f(x) = \pm \infty\) (depending on the sign), but it is helpful to note that, by the definition of the Limit of a Function, this limit does not exist, since \(f(x)\) is not arbitrarily close to a single real number.
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We know, from our adage “Fractions with small denominators are big, and fractions with big denominators are small,” that this type of behavior happens when the denominator is tiny compared to the numerator. We can summarize this:
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These kinds of limits are great, in that they’re consistent to identify. We just look for this tiny denominator compared to the numerator, and go from there. We also know a lot about these types of limits, and can summarize this below.
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Evaluating Infinite Limits.

Once we know that \(\displaystyle \lim_{x\to a}f(x) \stackrel{?}{\to} \frac{\#}{0}\text{,}\) we know a bunch of information right away!
  • This limit doesn’t exist.
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  • The function \(f(x)\) has a vertical asymptote at \(x=a\text{,}\) causing these unbounded \(y\)-values near \(x=a\text{.}\)
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  • The one sided limits must be either \(\infty\) or \(-\infty\text{.}\)
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  • We only need to focus on the sign of the one sided limits! And signs of products and quotients are easy to follow.
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So a pretty typical process is to factor as much as we can, and check the sign of each factor (in a numerator or denominator) as \(x\to a^-\) and \(x\to a^+\text{.}\) From there, we can find the sign of \(f(x)\) in both of those cases, which will tell us the one-sided limit.
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Example 1.4.3.

For each function, find the relevant one-sided limits at the input-value mentioned. If you can use a two-sided limit statement to discuss the behavior of the function around this input-value, then do so.
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(a)
\(\left(\dfrac{2x^2-5x+1}{x^2+8x+16}\right)\) and \(x=-4\)
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Solution.
Let’s first factor the denominator: we want to see the factor \((x+4)\text{,}\) where we divide by \(0\text{.}\)
\begin{equation*} \left(\dfrac{2x^2-5x+1}{x^2+8x+16}\right) = \left(\dfrac{2x^2-5x+1}{(x+4)^2}\right) \end{equation*}
Now, when \(x\to 4\text{,}\) we have \(\left(\dfrac{2x^2-5x+1}{(x+4)^2}\right)\stackrel{?}{\to} \frac{13}{0}\text{.}\) This is our \(\frac{\#}{0}\) form that tells us we have an infinite limit. We’re going to look at the one-sided limits, but let’s notice something:
  • For \(x\)-values close to \(4\text{,}\) we expect that the numerator will be close to \(13\text{.}\) No matter what “side” of \(13\) it is, it’s still going to be positive.
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  • For \(x\)-values close to \(4\text{,}\) we expect the denominator to be close to \(0\text{.}\) Depending on what side of \(0\) it is, it could be positive or negative.
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So in our one-sided limits, we know that these both will be infinite limits (the function values will either approach \(-\infty\) or \(\infty\)). The only difference in these is the sign. So let’s check the signs!
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First, we’ll consider \(x\to -4^-\text{.}\) That’s \(x\lt -4\text{.}\) We know the numerator is positive, since it is close to \(13\text{,}\) and we also know that the factor \(x+4\lt 0\text{.}\)
\begin{align*} \lim_{x\to-4^-} \left(\dfrac{2x^2-5x+1}{(x+4)^2}\right) \amp = \lim_{x\to-4^-} \left(\dfrac{\overbrace{2x^2-5x+1}^{\approx 13}}{(\underbrace{x+4}_{\approx 0})^2}\right) \\ \amp = \dfrac{\oplus}{(\ominus)^2}\\ \amp = \oplus \end{align*}
So we know that \(\displaystyle \lim_{x\to-4^-} \left(\dfrac{2x^2-5x+1}{(x+4)^2}\right) \) is an infinite limit, and we know that it is positive.
\begin{equation*} \lim_{x\to-4^-} \left(\dfrac{2x^2-5x+1}{(x+4)^2}\right) =\infty \end{equation*}
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Now we can check the signs of these factors when \(x\to-4^+\) (and so \(x\gt-4\)). The numerator is still close to \(13\text{,}\) and so still negative. Now, \((x+4)\) is positive, and is still being squared.
\begin{align*} \lim_{x\to-4^+} \left(\dfrac{2x^2-5x+1}{(x+4)^2}\right) \amp = \lim_{x\to-4^+} \left(\dfrac{\overbrace{2x^2-5x+1}^{\approx 13}}{(\underbrace{x+4}_{\approx 0})^2}\right) \\ \amp = \dfrac{\oplus}{(\oplus)^2}\\ \amp = \oplus \end{align*}
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Since this is an infinite limit and is also positive, we know that:
\begin{equation*} \lim_{x\to-4^+} \left(\dfrac{2x^2-5x+1}{(x+4)^2}\right) =\infty\text{.} \end{equation*}
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Since the function approaches \(\infty\) on both sides, we can say that
\begin{equation*} \lim_{x\to-4} \left(\dfrac{2x^2-5x+1}{(x+4)^2}\right) =\infty\text{,} \end{equation*}
but we know that this limit does not exist (since the function values are not close to a single real number).
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(b)
\(\left(\dfrac{4x^2-x^5}{x^2-4x+3}\right)\) and \(x=1\)
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Solution.
We can start this in a similar way: factor the denominator to see the places where we divide by \(0\text{:}\)
\begin{equation*} \left(\dfrac{4x^2-x^5}{x^2-4x+3}\right) = \left(\dfrac{4x^2-x^5}{(x-3)(x-1)}\right)\text{.} \end{equation*}
Now, when we think about the limit as \(x\to 1^-\text{,}\) we’re thinking about \(x\lt 1\text{.}\) We can check the signs, again! The numerator has \(4x^2-x^5 \to 3\text{,}\) and so for \(x\)-values slightly smaller than \(1\text{,}\) this numerator is close to \(3\text{,}\) and so positive. Similarly, \((x-3)\to -2\text{,}\) and so for \(x\)-values close to but less than \(1\text{,}\) this is negative. Then, we can see that \((x-1)\to 0\text{.}\) This is the part that gives us the \(\frac{\#}{0}\) form.
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For \(x\lt 1\text{,}\) we have \((x-1)\lt 0\text{,}\) and so this is negative.
\begin{equation*} \lim_{x\to 1^-}\left(\dfrac{\overbrace{4x^2-x^5}^{\oplus}}{(\underbrace{x-3}_{\ominus})(\underbrace{x-1}_{\ominus})}\right)\to \oplus \end{equation*}
And so this limit is a positive infinite limit:
\begin{equation*} \lim_{x\to 1^-} \left(\dfrac{4x^2-x^5}{(x-3)(x-1)}\right) = \infty\text{.} \end{equation*}
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Now we can similarly check the signs when \(x\to 1^+\text{,}\) which is when \(x\gt 1\text{.}\) Note that, since \(x\) is still close to \(1\text{,}\) the numerator and the factor \((x-3)\) will retain their sign. But, for the factor \((x-1)\text{,}\) we get \((x-1)\gt 0\text{.}\)
\begin{equation*} \lim_{x\to 1^-}\left(\dfrac{\overbrace{4x^2-x^5}^{\oplus}}{(\underbrace{x-3}_{\ominus})(\underbrace{x-1}_{\oplus})}\right)\to \ominus \end{equation*}
And so this limit is a negative infinite limit:
\begin{equation*} \lim_{x\to 1^+} \left(\dfrac{4x^2-x^5}{(x-3)(x-1)}\right) = -\infty\text{.} \end{equation*}
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(c)
\(\sec(\theta)\) and \(\theta = \dfrac{\pi}{2}\)
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Solution.
We can think of \(\sec(\theta)\) as a reciprocal: \(\dfrac{1}{\cos(\theta)}\text{.}\) Now, we can see that \(\cos\left(\frac{\pi}{2}\right) = 0\text{,}\) hence this is an infinite limit.
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Let’s visualize \(\cos\left(\frac{\pi}{2}\right)\text{,}\) so that we can tell the sign of this denominator when \(\theta\) is on either side of \(\frac{\pi}{2}\text{.}\)
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A unit circle with a point at (0,1) labeled (cos(pi/2),sin(pi/2)). On the right side of the point is a small blue arc labeled theta less than pi/2. On the left side of the point is a small red arc labeled theta greater than pi/2.
Figure 1.4.4.
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We can see that for \(\theta\lt\frac{\pi}{2}\text{,}\) we are looking at a point in the first quadrant with a positive horizontal component. So, in this case, \(\cos(\theta)\gt 0\text{.}\)
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For the case when \(\theta\gt\frac{\pi}{2}\text{,}\) though, we are looking in the second quadrant with a negative horizontal component. So we see that \(\cos(\theta)\lt 0\text{.}\)
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All of this to say:
\begin{equation*} \lim_{\theta\to\frac{\pi}{2}^-} \sec(\theta) = \infty \end{equation*}
\begin{equation*} \lim_{\theta\to\frac{\pi}{2}^+} \sec(\theta) = -\infty \end{equation*}
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These are limits where \(f(x)\to \pm\infty\) when \(x\to a\) where \(a\) is a real number. What about the other way around? Is there a case where \(x\to\pm\infty\text{,}\) and what would we be looking at in the behavior of \(f(x)\text{?}\)
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Subsection End Behavior Limits

Activity 1.4.2. What Happens When We Divide by Infinity?

Again, we need to start by making something clear: if we were really going to try divide some real number by infinity, then we would need to rebuild our definition of what it means to divide. In the context we’re in right now, we only have division defined as an operation for real (and maybe complex) numbers. Since infinity is neither, then we will not literally divide by infinity.
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When we talk about “dividing by infinity” here, we’re again talking about the behavior of some function in a limit. We want to consider what it might look like to have a function that involves division where the denominator gets arbitrarily large (positive or negative) (or, the limit of the denominator is infinite).
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(b)
Let’s again consider \(12\div N\text{.}\) What is the value of this division problem when:
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(c)
Consider a function \(f(x) = \frac{12}{x}\text{.}\) What happens to the value of this function when \(x\to \infty\text{?}\) Note that this means that the \(x\)-values we’re considering most are very large and positive.
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(d)
Consider a function \(f(x) = \frac{12}{x}\text{.}\) What happens to the value of this function when \(x\to -\infty\text{?}\) Note that this means that the \(x\)-values we’re considering most are very large and negative.
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(e)
Why is there no difference in the behavior of \(f(x)\) as \(x\to \infty\) compared to \(x\to -\infty\) when the sign of the function outputs are opposite (\(f(x)\gt 0\) when \(x\to \infty\) and \(f(x) \lt 0\) when \(x\to-\infty\))?
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Definition 1.4.5. Limit at Infinity.

If \(f(x)\) is defined for all large and positive \(x\)-values and \(f(x)\) gets arbitrarily close to the single real number \(L\) when \(x\) gets sufficiently large, then we say:
\begin{equation*} \lim_{x\to\infty} f(x) = L\text{.} \end{equation*}
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Similarly, if \(f(x)\) is defined for all large and negative \(x\)-values and \(f(x)\) gets arbitrarily close to the single real number \(L\) when \(x\) gets sufficiently negative, then we say:
\begin{equation*} \lim_{x\to-\infty} f(x) = L\text{.} \end{equation*}
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In the case that \(f(x)\) has a limit at infinity that exists, then we say \(f(x)\) has a horizontal asymptote at \(y=L\text{.}\)
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Lastly, if \(f(x)\) is defined for all large and positive (or negative) \(x\)-values and \(f(x)\) gets arbitrarily large and positive (or negative) when \(x\) gets sufficiently large (or negative), then we could say:
\begin{equation*} \lim_{x\to-\infty} f(x) = \pm\infty \text{ or } \lim_{x\to\infty} f(x) = \pm\infty\text{.} \end{equation*}
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Because the primary focus for limits at infinity is the end behavior of a function, we will often refer to these limits as end behavior limits.
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We’re going to think about end-behavior of many types of functions, but we want to start with some small examples and build from there. So we’re going to start by thinking about power functions. We’re going to think about power functions in two ways:
  1. Power functions in the form of \(x^p\) (with \(p\gt 0\)). It shouldn’t be too much work to convince yourself that these functions always have limits where, as \(x\to \pm\infty\text{,}\) \(x^p\to\pm\infty\) as well.
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  2. Reciprocal power functions, in the form \(\dfrac{1}{x^p}\) (still with \(p\gt 0\)). We already should have an idea of what’s going to happen in these, based on Activity 1.4.2.
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Our last result is one that you might already know, but we’ll provide some more justification for this. We can use our knowledge of end-behavior limits for both type of power functions to think about the end-behavior limits of polynomials in general!
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Proof.

Consider the polynomial function:
\begin{equation*} p(x) = a_n x^n + a_{n-1}x^{n-1} + ... + a_2 x^2 + a_1 x + a_0 \end{equation*}
where \(n\) is some integer and \(a_k\) is a real number for \(k=0,1,2,...,n\text{.}\) For simplicity, we will consider only the limit as \(x\to\infty\text{,}\) but we could easily repeat this exact proof for the case where \(x\to-\infty\text{.}\)
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Before we consider this limit, we can factor out \(x^n\text{,}\) the variable with the highest exponent:
\begin{align*} p(x) \amp =a_n x^n + a_{n-1}x^{n-1} + ... + a_2 x^2 + a_1 x + a_0\\ \amp = x^n \left( \frac{a^nx^n}{x^n} + \frac{a_{n-1}x^{n-1}}{x^n} + ... + \frac{a_2x^2}{x^n} + \frac{a_1x}{x^n} + \frac{a_0}{x^n}\right)\\ \amp= x^n \left( a_n + \frac{a_{n-1}}{x} + ... + \frac{a_2}{x^{n-2}} + \frac{a_1}{x^{n-1}} + \frac{a_0}{x^n} \right) \end{align*}
Now consider the limit of this product:
\begin{align*} \lim_{x\to\infty} p(x) \amp = \lim_{x\to\infty} x^n \left( a_n + \frac{a_{n-1}}{x} + ... + \frac{a_2}{x^{n-2}} + \frac{a_1}{x^{n-1}} + \frac{a_0}{x^n} \right) \\ \amp =\left(\lim_{x\to\infty} x^n\right)\left(\lim_{x\to\infty} a_n + \frac{a_{n-1}}{x} + ... + \frac{a_2}{x^{n-2}} + \frac{a_1}{x^{n-1}} + \frac{a_0}{x^n} \right) \end{align*}
We can see that in the second limit, we have a single constant term, \(a_n\text{,}\) followed by reciprocal power functions. Then, due to Theorem 1.4.6, we know that the second limit will by \(a_n\text{,}\) since the reciprocal power functions will all approach 0.
\begin{align*} \lim_{x\to\infty} p(x) \amp =\left(\lim_{x\to\infty} x^n\right)\left(\lim_{x\to\infty} a_n + \frac{a_{n-1}}{x} + ... + \frac{a_2}{x^{n-2}} + \frac{a_1}{x^{n-1}} + \frac{a_0}{x^n} \right)\\ \amp = \left(\lim_{x\to\infty} x^n\right)\left(a_n + 0 + ... + 0 + 0 + 0\right)\\ \amp = \left(\lim_{x\to\infty} x^n\right)\left(a_n\right)\\ \amp = \lim_{x\to\infty} a_n x^n \end{align*}
And so \(\displaystyle \lim_{x\to\infty} p(x) = \lim_{x\to\infty} a_n x^n\) as we claimed.
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Instead of spending our time thinking about these polynomial end-behavior limits too specifically (since we might already be familiar with this result), and just focus on using these polynomials in the middle of larger problems.
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Example 1.4.8.

For each function, find the limits as \(x\to \infty\) and \(x\to-\infty\text{.}\)
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(a)
\(\left(\dfrac{2x^2-5x+1}{x^2+8x+16}\right)\)
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Hint 1.
You can think about the limit in the numerator and the limit in the denominator. Based on Theorem 1.4.7, which terms will be the ones to dictate the behavior of the numerator and denominator?
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Hint 2.
What happens when you think about just those dominant terms in the numerator and denominator and reduce the fraction? What is left?
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Solution.
We’ll start with the same kind of factoring that is used in the proof of Theorem 1.4.7.
\begin{align*} \left(\dfrac{2x^2-5x+1}{x^2+8x+16}\right) \amp = \left(\dfrac{2x^2}{x^2}\right)\left(\dfrac{1-\frac{5}{2x}+\frac{1}{2x^2}}{1+\frac{8}{x}+\frac{16}{x^2}}\right) \\ \amp = 2\left(\dfrac{1-\frac{5}{2x}+\frac{1}{2x^2}}{1+\frac{8}{x}+\frac{16}{x^2}}\right) \end{align*}
Now we can apply the limits as \(x\to\infty\) and \(x\to-\infty\text{,}\) since the reciprocal power functions will all \(\to 0\text{.}\)
\begin{align*} \lim_{x\to\infty} 2\left(\dfrac{1-\frac{5}{2x}+\frac{1}{2x^2}}{1+\frac{8}{x}+\frac{16}{x^2}}\right) \amp = 2(1) = 2 \\ \lim_{x\to-\infty} 2\left(\dfrac{1-\frac{5}{2x}+\frac{1}{2x^2}}{1+\frac{8}{x}+\frac{16}{x^2}}\right) \amp = 2(1) = 2 \end{align*}
Alternatively, we could have done the following:
\begin{align*} \lim_{x\to \infty} \left(\dfrac{2x^2-5x+1}{x^2+8x+16}\right) \amp = \dfrac{\lim_{x\to\infty}(2x^2-5x+1)}{\lim_{x\to\infty}(x^2+8x+16)}\\ \amp = \frac{\lim_{x\to\infty} 2x^2}{\lim_{x\to\infty}x^2}\\ \amp = \lim_{x\to\infty} \frac{2x^2}{x^2}\\ \amp = \lim_{x\to\infty} 2 = 2 \end{align*}
The same process could be used to show that:
\begin{equation*} lim_{x\to -\infty} \left(\dfrac{2x^2-5x+1}{x^2+8x+16}\right)=2\text{.} \end{equation*}
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(b)
\(\left(\dfrac{4x^2-x^5}{x^2-4x+3}\right)\)
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Hint.
Be careful about which term (in the numerator) will persist!
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(c)
\(\dfrac{|x|}{3x}\)
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Hint.
It might be helpful to remind yourself of the definition of the absolute value function:
\begin{equation*} |x| = \begin{cases} -x \amp \text{when } x\lt0\\ x \amp \text{when } x\geq 0 \end{cases}\text{.} \end{equation*}
This means you can replace \(|x|\) with either \(x\) or \(-x\) in the limits as \(x\to\infty\) and \(x\to-\infty\text{.}\)
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Practice Problems Practice Problems

1.

Use the graph of the function \(f(x)\) below to determine each limit.
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A rational function with a horizontal asymptote at y=0. At x=-3, the function increases up to a vertical asymptote and then decreases down from it. At x=5, the function increases up to a vertical asymptote and then increases away from it. At x=10 the function decreases towards a vertical asymptote and then decreases away from it.
Figure 1.4.9. The function \(f(x)\text{.}\)
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(a)
\(\displaystyle\lim_{x\to-3^-} f(x)\)
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(b)
\(\displaystyle\lim_{x\to-3^+} f(x)\)
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(c)
\(\displaystyle\lim_{x\to-3} f(x)\)
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(d)
\(\displaystyle\lim_{x\to5^-} f(x)\)
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(e)
\(\displaystyle\lim_{x\to5^+} f(x)\)
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(f)
\(\displaystyle\lim_{x\to5} f(x)\)
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(g)
\(\displaystyle\lim_{x\to10^-} f(x)\)
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(h)
\(\displaystyle\lim_{x\to10^+} f(x)\)
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(i)
\(\displaystyle\lim_{x\to10} f(x)\)
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2.

Use the graph of the function \(g(x)\) below to determine each limit.
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A graph of a function increasing away from a horizontal asymptote at x=-2 towards (0,0), and then decreasing towards a vertical asymptote at x=1. It decreases away from the vertical asymptote at x=1 towards a horizontal asymptote ay y=2.
Figure 1.4.10. The function \(g(x)\text{.}\)
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(a)
\(\displaystyle\lim_{x\to-\infty}g(x)\)
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(b)
\(\displaystyle\lim_{x\to\infty}g(x)\)
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3.

Use the graph of the function \(h(x)\) below to determine each limit.
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A function increases from the bottom left corner of the graph towards an open point at (-1,0). Then it decreases away from a point at (-1,1) towards a horizontal asymptote at y=0.
Figure 1.4.11. The function \(h(x)\text{.}\)
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(a)
\(\displaystyle\lim_{x\to-\infty}h(x)\)
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(b)
\(\displaystyle\lim_{x\to\infty}h(x)\)
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4.

For each limit, fill in the table in a way that determines the limit, and report your answer.
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(a)
\(\displaystyle\lim_{x\to4^-} \left(\frac{3x}{x-4}\right)\)
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\(x\) \(4\)
\(\frac{3x}{x-4}\) does not exist
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(b)
\(\displaystyle\lim_{x\to4^+} \left(\frac{3x}{x-4}\right)\)
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\(x\) \(4\)
\(\frac{3x}{x-4}\) does not exist
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(c)
\(\displaystyle\lim_{x\to 1} \left(\frac{x-3}{(2x-4)(x-1)^2}\right)\)
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\(x\) \(1\)
\(\frac{x-3}{(2x-4)(x-1)^2}\) does not exist
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5.

Evaluate the following limits analytically. Also evaluate the relevant one-sided limits when needed.
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(a)
\(\displaystyle\lim_{x\to-\infty} \left(2+4x-5x^2+\dfrac{x^3}{7} - \dfrac{8x^4}{9}\right)\)
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(b)
\(\displaystyle\lim_{x\to\infty} \left( \dfrac{5x^3-4x+3}{2x+6}\right)\)
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(c)
\(\displaystyle\lim_{x\to -3^-} \left( \dfrac{5x^3-4x+3}{2x+6}\right)\)
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(d)
\(\displaystyle\lim_{x\to-\infty} \left( \dfrac{x-10}{3x+4}\right)\)
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(e)
\(\displaystyle\lim_{x\to 2} \left( \dfrac{4x-1}{x^4-16}\right)\)
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(f)
\(\displaystyle\lim_{x\to\infty} \left( \dfrac{6-5x^5}{2+3x^2}\right)\)
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(g)
\(\displaystyle\lim_{x\to 1} \left( \dfrac{x^2+1}{x^3-2x^2+x}\right)\)
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(h)
\(\displaystyle\lim_{\theta\to3\pi/2} \sin(\theta)\tan(\theta)\)
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6.

For each of the following functions, find the locations of any vertical asymptotes and report the behavior of the functions around those asymptotes using limit statements.
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(a)
\(f(x) = \dfrac{|x-4|}{x+2}\)
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(b)
\(g(x) = \dfrac{\sin x}{x^2-1}\)
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(c)
\(h(t) = \dfrac{4}{t} + \dfrac{9}{t-1}\)
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Hint.
If you’d rather, you can add the terms together to get:
\begin{equation*} h(t)=\frac{13t-4}{t(t-1)}\text{.} \end{equation*}
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(d)
\(s(w) = \dfrac{w^3-3w^2+w-1}{w^2-5w+6}\)
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(e)
\(f(x) = \dfrac{e^x}{(x-3)^4}\)
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7.

For each of the following functions, find the locations of any horizontal asymptotes using limit statements (as \(x\to\infty\) and \(x\to-\infty\)).
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(a)
\(f(x)= 2 + \dfrac{10}{x^2}\)
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(b)
\(g(x)=\dfrac{4x^5-3x^3+5x-1}{10x^2+5x^3-5x^4}\)
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(c)
\(h(x) = \dfrac{3x^2+4x+\sqrt{x}}{1-\sqrt{x}-2x^2}\)
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(d)
\(j(x) = \dfrac{3}{5 + 10^{1/x}}\)
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