§1.2 The notion of limit

Introduction

The Notion of Limit

Instantaneous Velocity

Summary

A function $g$ satisfies $g(x) = 4 - x^2$ on $-2.5 \lt x \le 1$, except that the point $(0,4)$ is an open circle and the filled point $(0,1)$ sets $g(0) = 1$. On $1 \lt x \lt 3.5$ it follows $g(x) = 3 - x$, with only an open circle at $(2,1)$. Which statement is true?

For the same $g$ ($g(x) = 4 - x^2$ on $-2.5 \lt x \le 1$ with $g(0) = 1$, and $g(x) = 3 - x$ on $1 \lt x \lt 3.5$ with a hole at $(2,1)$), which list correctly describes $\lim_{x \to -1} g(x)$, $\lim_{x \to 0} g(x)$, $\lim_{x \to 1} g(x)$, $\lim_{x \to 2} g(x)$?

Investigate $f(x) = \frac{4-x^2}{x+2}$ near $a = -2$ and $g(x) = \sin\left(\frac{\pi}{x}\right)$ near $a = 0$. What do numerical and algebraic approaches show?

Evaluate $\displaystyle \lim_{x \to 1} \frac{x^2 - 1}{x - 1}$.

Evaluate $\displaystyle \lim_{x \to 0} \frac{(2+x)^3 - 8}{x}$.

Evaluate $\displaystyle \lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x}$. Give the exact value as a decimal or fraction.

For $s(t) = t^2$, the average velocity on $[3, 3+h]$ simplifies exactly to $6 + h$ for every $h$ with $h \ne 0$.

For $s(t) = t^2$ meters with $t$ in minutes, the average velocity on $[3, 3+h]$ simplifies to $6 + h$. Use this to compute the average velocity on $[3, 3.2]$, in meters per minute.

For an object with position $s(t) = t^2$ meters at time $t$ minutes, the instantaneous velocity at $t = 3$ is $6$ meters per minute.

An object's position readings are $s(1.5) = 1$, $s(2.5) = 3$, $s(0.5) = 1$, and $s(1) = 1$ meters. Compute the average velocity on the interval $[1.5, 2.5]$, in meters per second.

An object moves along a line with position $s(t) = t^2$ meters at time $t$ minutes. Which expression gives the instantaneous velocity at $t = 3$?

From a position graph, an object has $s(2) = 2$, and the average velocity on $[1.5, 2.5]$ is $2$; a secant through $(2, 2)$ and a nearby point at $t = 2.1$ has slope about $2.5$. How does the instantaneous velocity at $t = 2$ compare with the average velocity on $[1.5, 2.5]$?

Let $f(x) = \frac{16 - x^4}{x^2 - 4}$. Evaluate $\displaystyle \lim_{x \to 2} f(x)$, giving the exact value.

Let $g(x) = -\frac{|x+3|}{x+3}$. What can be said about $\displaystyle \lim_{x \to -3} g(x)$?

It is possible to sketch the graph of a function with $f(2) = 1$ for which $\lim_{x \to 2} f(x)$ does not exist.

For a bungee jumper with height $s(t) = 100\cos(0.75t)\,e^{-0.2t} + 100$ feet, a spreadsheet evaluation of $\displaystyle \lim_{h \to 0} \frac{s(1+h) - s(1)}{h}$ gives about $-53.837$. What does this number mean?