CALCULUS · BEFORE DIFFERENTIATION

Limits: approaching & growth

Suppose a car travels 100 km in 2 hours. Its average speed is $100 \div 2 = 50\ \mathrm{km/h}$ , although the speedometer need not stay at 50 km/h. To describe its speed at one instant, we first take a short time interval containing that instant and divide the distance travelled by the elapsed time. We cannot put zero straight into the denominator. Instead, we make the interval shorter and ask whether the average speeds approach one value. When that limit exists, it gives the speed at the instant.

Keep the car question in mind while we try a smaller one. A function is a rule giving a result for each allowed input; a point on its graph records one input and its result. We want to know where the results head as the input gets closer to a particular number. You might ask, “Why not just put that number in?” Sometimes the value there does not exist, although nearby values still follow a pattern we can examine. The table and graph show these together. Approaching from the left and the right, we can check whether both sides head towards the same result, and whether the value at the point agrees with it.

01What do the left and right sides approach?

Play or scrub the approach. The two sample inputs never equal a. Switch cases to compare agreement, a hole, a jump, oscillation and unbounded growth.

f(x)=x^2

● From the left● From the rightArrow at edge = point continues beyond the view

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What value does the function approach near a?

We write $x\to a$ for “x gets closer and closer to a,” and $x\ne a$ for “look near a, but never at a itself.” The limit asks which value $f(x)$ approaches as $x$ does this. A left limit uses $x<a$ ; a right limit uses $x>a$ . + and − in the limit notation mean sides, not the sign of the answer.

x_L=1-h,\quad x_R=1+h

h=1.2

Side	x	f(x)
Left	-0.2	0.04
Right	2.2	4.84

\lim_{x\to1^-}f(x)=1

\lim_{x\to1^+}f(x)=1

\lim_{x\to1}x^2=1=f(1)

Write $x=1+h$ , where $h$ may be positive or negative. Then $x^2=(1+h)^2=1+2h+h^2$ , so its difference from 1 is $2h+h^2$ . When $\lvert h\rvert <1$ , the absolute difference is at most $2\lvert h\rvert+h^2\le 3\lvert h\rvert$ . By choosing $\lvert h\rvert$ small enough, we can make this smaller than any chosen positive number. Thus both sides approach 1. Since $f(1)=1$ too, this example is continuous at 1.

DNE means “does not exist.” A finite two-sided limit exists exactly when both one-sided limits exist and equal the same finite number. A finite animation illustrates the behavior; the formulas explain why it persists arbitrarily close to a.

02Which term eventually dominates?

We now let the input grow without bound, rather than approach one finite number. A power such as $x^3$ has a fixed exponent, here 3. The lower-degree terms may still grow, but they shrink relative to the leading term. Even a large coefficient such as 400 can delay dominance without changing the eventual limit. The graph also contains $e^x$ and $\ln x$ . Here $e\approx2.718$ is a fixed positive number: increasing the exponent of $e^x$ by 1 multiplies its value by $e$ . For positive $x$ , $\ln x$ asks which exponent $t$ gives $e^t=x$ ; it reverses that exponential rule. We will construct the exponential and justify the comparisons after inspecting the graph.

x	$2x^3$	$-2x^2$	1	$\frac{\|-2x^2\|}{\|2x^3\|}$
10	2000	-200	1	0.1
100	2.000e+6	-20000	1	0.01
1000	2.000e+9	-2.000e+6	1	0.001
10000	2.000e+12	-2.000e+8	1	0.0001
1.000e+6	2.000e+18	-2.000e+12	1	1.000e-6

The quadratic term does not tend to zero: its magnitude grows. Its ratio to the cubic term tends to zero. That is the precise meaning of “negligible” here.

e^x

x^2+1

x

\sqrt{x-1}

\ln x

Right edge of x-axis6

All curves use $x>0$ , but $\sqrt{x-1}$ only exists for $x\ge 1$ in the real numbers. On a finite interval a larger slope does not establish a higher asymptotic growth order.

For example, $\sqrt{x-1}$ becomes extremely steep near $x=1$ , yet it eventually grows much more slowly than $x$ . Compare ratios as $x\to \infty$ instead.

\ln x\ll\cdots\ll x^{1/3}\ll x^{1/2}\ll x\ll x^2\ll x^3\ll\cdots\ll e^x

Here $f \ll g$ means $f/g\to 0$ as $x\to \infty$ . Each exponent is a fixed positive number. This is an eventual growth hierarchy, not a pointwise inequality for every positive $x$ . The two sides have a useful inverse-function pattern: exponential ↔ logarithm and powers ↔ roots, with suitable domains.

The graph can suggest an order, but it only shows a finite interval. To justify the whole chain, we need to show that each smaller expression divided by the next one tends to zero. Start with powers. If $q>p>0$ , cancelling the common power leaves a positive power in the denominator.

\frac{x^p}{x^q}=\frac1{x^{q-p}}\longrightarrow0\qquad(x\to\infty)

This includes roots: $\sqrt{x}$ is x to the power $\tfrac12$ . For the shifted root in the graph, $0\le\frac{\sqrt{x-1}}{x}\le\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$ when $x\ge1$ , so it also becomes small compared with x. The upper bound can be made smaller than any chosen positive number by taking x large enough.

Why does an exponential eventually exceed every fixed power? Here is a construction that makes the reason visible in the algebra. For $t\ge0$ , we can define $e^t$ as the limit of $\left(1+t/N\right)^N$ as the positive integer N increases. In this product, choosing $t/N$ from exactly k of its N brackets gives the term below. There are $\frac{N(N-1)\cdots(N-k+1)}{k!}$ ways to make that choice; $k!$ means $1\times2\times\cdots\times k$ and removes the repeated orders of the same selection.

\begin{gathered}\left(1+\frac tN\right)^N\\\ge\frac{N(N-1)\cdots(N-k+1)}{k!}\frac{t^k}{N^k}\\\longrightarrow\frac{t^k}{k!}\end{gathered}

Every term is nonnegative, so keeping one term gives a lower bound. With k fixed, each factor $(N-j)/N$ approaches 1. Hence $e^t\ge t^k/k!$ . To compare $x^p$ with $e^x$ , choose an integer k greater than p. We now have a bound whose limit we already understand.

0\le\frac{x^p}{e^x}\le\frac{k!}{x^{k-p}}\longrightarrow0

Finally, $\ln x$ is the number t for which $x=e^t$ . This substitution is useful because it turns the logarithm into t and the power $x^p$ into $e^{pt}$ . Use the same bound with $k=2$ and input pt. For every fixed $p>0$ , the resulting upper bound tends to zero. These arguments justify eventual ratios; they do not say which graph is higher at every small input.

\begin{gathered}t=\ln x\longrightarrow\infty\\0\le\frac{\ln x}{x^p}=\frac{t}{e^{pt}}\\\le\frac{2}{p^2t}\longrightarrow0\end{gathered}

03Watch the growth gap open up.

Keep every original term. The cubic terms grow faster than the quadratic terms, just as in the growth-rate comparison above. Increase x and compare the actual values in the numerator and denominator.

R(x)=\frac{2x^3-2x^2+1}{4x^3+400x^2-1000}

x $10$

Full expression R(x)0.0418837

Cubic terms alone:

2x^3 / 4x^3

0.5

One shared linear scale · bar length = absolute valueFull width =

40,000

NUMERATOR

2x^3

2,000

-2x^2

-200

1

1

DENOMINATOR

4x^3

4,000

400x^2

40,000

-1000

-1,000

The ruler zooms out as the numbers grow; it is shared by all six bars. Read the actual values on the right: the quadratic terms are growing too. Their bars fall behind because the cubic terms grow faster. A tiny dot marks a nonzero term below one pixel; minus signs remain in the values.

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At $x=10$ , the factor 400 makes the quadratic term in the denominator much larger than its cubic term. The limit describes what happens eventually, not what happens at every $x$ .

\text{large }x:\quad\frac{2x^3-2x^2+1}{4x^3+400x^2-1000}\approx\frac{2x^3}{4x^3}=\frac12

\boxed{\lim_{x\to\infty}\frac{2x^3-2x^2+1}{4x^3+400x^2-1000}=\frac12}

Both leading terms are cubic, so they grow at the same order; their coefficient ratio is 2:4. The lower-growth terms become negligible by comparison. The approximation sign matters: at a finite $x$ , the full expression has not become exactly $1/2$ .

Let us turn the comparison into an exact calculation. The largest power is $x^3$ . Divide every term of both numerator and denominator by $x^3$ ; dividing both by the same nonzero number leaves the fraction unchanged. We use $x>0$ because we are taking x towards positive infinity.

\begin{gathered}\frac{2x^3-2x^2+1}{4x^3+400x^2-1000}\\=\frac{2-2/x+1/x^3}{4+400/x-1000/x^3}\end{gathered}

For example, $400x^2/x^3=400/x$ : only two factors of x cancel, leaving one in the denominator. As x increases without bound, $1/x$ and $1/x^3$ tend to zero. Multiplying them by fixed numbers does not change that limit. The numerator approaches 2, while the denominator approaches 4, which is not zero. We can now take their ratio.

\begin{gathered}\lim_{x\to\infty}\frac{2-2/x+1/x^3}{4+400/x-1000/x^3}\\=\frac{2-0+0}{4+0-0}\\=\frac12\end{gathered}

Teaching sequence informed by the supplied Calculus from Scratch manuscript, Limits chapter, sections on one-sided limits and polynomial/rational limits. Terminology distinguishes finite limits from infinite limits.