Calculus I · Limits and Continuity · lesson

Why Limits Matter in Calculus

Concept

Learning objectives

Explain the difference between average and instantaneous change; compute average rates on shrinking intervals; interpret a limit as a number approached by those rates.

What a Limit Means

The Problem Calculus Is Trying to Solve

Suppose you walk 12 feet in 3 seconds. Your average speed is

\frac{12\text{ feet}}{3\text{ seconds}}=4\text{ feet per second}.

Nothing mysterious has happened. Two different times were compared, so the elapsed time in the denominator was not zero.

Now ask a harder question: how fast were you moving at exactly $t=2$ seconds? A single instant has no duration. If we try to compare the position at $t=2$ with itself, we get

\frac{\text{position at }2-\text{position at }2}{2-2} =\frac00,

which is undefined.

Calculus solves this problem by refusing to measure over an interval of zero length. Instead, it measures over ordinary intervals and makes those intervals shorter and shorter. If the resulting average rates settle toward one number, that number is the instantaneous rate.

Concept

Imagine a video of a moving bicycle. One frame does not show speed. But if you compare two frames that are very close together, you can estimate the speed. Use frames closer and closer together, and the estimates may settle toward the speed at the chosen moment. A limit records the number those estimates approach.

Average rate of change

For a function $f$ , the average rate of change from $x=a$ to $x=b$ is

\boxed{\frac{f(b)-f(a)}{b-a}}.

This is also the slope of the secant line through the points

(a,f(a))\qquad\text{and}\qquad(b,f(b)).

Guided walkthrough

A straight-moving cart

A toy cart has position

s(t)=3t+1

in feet after $t$ seconds. Find its average velocity from $t=2$ to $t=5$ .

Show worked solution

First find the two positions:

s(2)=3(2)+1=7,

s(5)=3(5)+1=16.

The position changed by

16-7=9\text{ feet}.

The time changed by

5-2=3\text{ seconds}.

Therefore,

\frac{s(5)-s(2)}{5-2} =\frac{9}{3} =\boxed{3\text{ ft/s}}.

The cart moves at a constant rate, so every average velocity is $3$ ft/s.

Worked example

A ball thrown upward

A ball has height

s(t)=64t-16t^2,

where $s$ is measured in feet and $t$ in seconds. Find the average velocity from $t=1$ to $t=2$ .

Show worked solution

Find the starting height.

s(1)=64(1)-16(1)^2=64-16=48.

Find the ending height.

s(2)=64(2)-16(2)^2=128-64=64.

Divide the change in height by the change in time.

\frac{s(2)-s(1)}{2-1} =\frac{64-48}{1} =\boxed{16\text{ ft/s}}.

The positive sign says that, on average, the ball moved upward over this interval.

Shrinking the interval

To estimate the velocity at exactly $t=1$ , compare $t=1$ with $t=1+h$ , where $h$ is a small nonzero number. The average velocity is

\frac{s(1+h)-s(1)}{h}.

Now simplify very carefully:

\begin{aligned} \frac{s(1+h)-s(1)}{h} &=\frac{64(1+h)-16(1+h)^2-48}{h}\\ &=\frac{64+64h-16(1+2h+h^2)-48}{h}\\ &=\frac{64+64h-16-32h-16h^2-48}{h}\\ &=\frac{32h-16h^2}{h}\\ &=\frac{h(32-16h)}{h}\\ &=32-16h,\qquad h\ne0. \end{aligned}

Notice why $h\ne0$ matters. We may cancel $h$ because the shrinking intervals always have nonzero length. We are studying what happens as $h$ gets close to zero, not substituting zero before simplifying.

Reference table
$h$	Average velocity $32-16h$
$1$	$16$
$0.5$	$24$
$0.1$	$30.4$
$0.01$	$31.84$
$0.001$	$31.984$
$-0.001$	$32.016$
$-0.01$	$32.16$

The values approach $32$ . We write

\boxed{\lim_{h\to0}\frac{s(1+h)-s(1)}{h}=32}.

The instantaneous velocity at $t=1$ is $32$ ft/s.

Interactive graph of secant lines approaching the tangent to a position curve. — **Read the graph**
Secant lines through $P=(1,48)$ approach the tangent line as the second point moves toward $P$ .

Interactive graph exploration

Function: $s(t)=64t-16t^2$ . Window: $0\le t\le4$ , $0\le s\le72$ . Fix $P=(1,48)$ . Allow a movable point $Q=(1+h,s(1+h))$ with controls for positive and negative $h$ . Display the secant slope $32-16h$ and the tangent line $s-48=32(t-1)$ . Include units in all labels.

Common mistake

The instantaneous rate is not obtained by pretending $0/0=0$ , $1$ , or infinity. The expression $0/0$ tells us that direct substitution has failed. The limit process asks what the quotient approaches for nonzero values that become arbitrarily small.

Quick check

A particle has position $p(t)=5t^2$ . Find its average velocity from $t=2$ to $t=2+h$ , simplify, and predict the instantaneous velocity at $t=2$ .

Answer.

\frac{5(2+h)^2-20}{h} =\frac{20h+5h^2}{h} =20+5h,

so the instantaneous velocity is $20$ .

After the explanation

Use the section idea

Reading lens

What are nearby outputs doing as the input approaches the target from both sides?

Mental model

Imagine tightening a window around the target input and watching where all nearby outputs are forced to gather.

Decision

Read the left-hand and right-hand behavior separately first; combine them only after both sides approach the same output.

Common trap

The function value at the target can be missing or deliberately moved, so never substitute a plotted dot for evidence from both sides.

Check yourself

You understand the section when you can explain a limit from a graph, table, and sentence without confusing it with the function value.

Source & rights

Original instruction with traceable references.

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