Section 1.7
Table of Contents
Formal definition of the limit
Neighborhoods #

What numbers are within 2 units of 1?
Write this as an interval and try completing this inequality: $ x  ?  < 2$Click to reveal the answer.
$(1, 3)$. We'll write this as an inequality $x1 < 2$ 
What numbers are within 1/2 of a unit of 1?
Write this as an interval and try completing this inequality: $ x  ?  < 1/2$Click to reveal the answer.
$\left(\frac12, \frac32\right)$. We'll write this as an inequality $x1 < 1/2$
The idea #
 You’re given a function and a particular $x$ value you’re interested in (this value is $a$).
 Pick a tiny number. 5? Sure. 0.5? Nice. 0.000005? Why not?
 This number is $\epsilon$.
 This forms an “epsilon neighborhood” of $y$values.
 Now the challenge is to find a corresponding number (most likely similarly tiny):
 This number is $\delta$
 This forms a “delta neighborhood” of $x$values.
 The limit of the function is $L$ if the function’s $y$values are always inside the epsilon neighborhood when the $x$values are in the delta neighborhood (the function always stays within the delta/epsilon rectangle in the picture):
Exploring the idea #
Here’s a Geogebra activity to explore the $\delta\epsilon$ definition of a limit for a function: https://www.geogebra.org/m/QgtVVage
and here’s a video of me describing how this works and our goal.
Formula Definition of the limit #
Let $f$ be a function defined on an open interval containing $a$. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write $$ \lim_{x\to a} f(x) = L $$ if for every $\epsilon > 0$ there is a corresponding number $\delta > 0$ so that if $$ 0 < x  a < \delta \text{ then } f(x)  L < \epsilon $$
Example #
Prove that $\displaystyle \lim_{x\to 4} \left(1 + \frac12 x\right) = 3$
Here’s the formal solution accompanying the work in the video
Click to reveal the answer.
Given $\epsilon > 0$, choose $\delta = 2\epsilon$. If $0 < x4 < \delta$, then $$ \left1+\frac12x 3\right = \left\frac12 x  2\right = \frac12 x  4 < \frac12 \cdot \delta = \frac12 \cdot 2\epsilon = \epsilon $$ Thus if $0 < x4 < \delta$, then $(1 + \frac12x)  4 < \epsilon$. Therefore, by the definition of the limit, $$ \lim_{x\to 4} \left(1 + \frac12 x\right) = 3 $$Now you try #
Prove that $\displaystyle \lim_{x\to 1} \left(2 + 5x\right) = 7$

First, start your “scratch work” and find a value of $\delta$ that will work for any $\epsilon > 0$.

Now your actual solution starts with the text
Let let $\epsilon >0$ be given, and choose $\delta = $ [your value]. If $0 < x1 < \delta$, then …
and continue like the above example’s solution.
Example #
Prove that $\displaystyle \lim_{x\to 0} x^2 = 0$
Note: Along the way, we’ll use the fact that $\sqrt{a^2} = a$ for all real numbers $a$.
Now practice! #
Head over to WebAssign and work on section 1.7. If you have questions, let me know!