Limit by epsilon-delta proof: Example 1

We have discussed extensively the meaning of the $\epsilon-\delta$ definition.  In this post, we are going to learn some strategies to prove limits of functions by definition.  The meat of the proof is finding a suitable $\delta$ for all possible $\epsilon$ values.

Recall that the definition states that the limit of $f(x) = L$ as $x$ approaches $a$, if for all $\epsilon > 0$, however small, there exists a $\delta > 0$ such that if $0 < | x - a| < \delta$, then $|f(x) - L| < \epsilon$.

Example 1: Let $f(x) = 3x + 5$.  Prove that $\lim_{x \to 2} f(x) = 11$

If we are going to study definition limit above, and apply it to the given function, we have $\lim_{x \to 2} 3x + 5 = 11$, if for all $\epsilon > 0$, however small, there exists a $\delta > 0$ such that if  $0 < |x - 2| < \delta$, then $|3x + 5 - 11| < \epsilon$.  We want to find the value of $\delta$, in terms of $\epsilon$; therefore, we can manipulate one of the inequalities to the other’s form.  In particular, we will manipulate $|3x + 5 - 11| < \epsilon$ to an expression such that the expression inside the absolute value sign will become $x - 2$.

Simplifying, we have $|3x - 6| = 3|x-2|< \epsilon$.  This expression is equivalent to  $3|x-2|< \epsilon$. Dividing both sides by $3$, we have $|x-2|< \frac{\epsilon}{3}$. Now, both inequalities have the same form.

That means that whatever value of $\epsilon$ is, we can find a $\delta = \frac{\epsilon}{3}$ satisfying the conditions above.  To explain further, let us have a specific example.

Let us choose a small $\epsilon = 0.6$. From the definition, for $\epsilon = 0.6$, there exists a $\delta=\frac{\epsilon}{3} =\frac{0.6}{3}= 0.2$  such that if  $0 < | x - 2| < 0.2$, then $|3x + 5 - 11| < 0.6$. The geometric interpretation of this statement is shown below.

Click the picture to view GeoGebra applet by Sylvain Bérubé

Now, let us intepret the definition. While reading these statements, look at the third diagram above:

1. Our $L=11$, we have chosen $\epsilon = 0.6$, and our computation gives us $\delta = \frac{\epsilon}{3} = \frac{0.6}{3} =0.2$.
2. The statement, let $\epsilon = 0.6$ means that we are creating an interval which is $(L - \epsilon, L + \epsilon) =(11-0.6,11+0.6)= (10.4,11.6)$.  This can be seen in the y-axis.
3. The statement $0 < |x - 2| < 0.2$  means that we are creating an interval $(a - \delta, a + \delta)=(2 -0.2,2+0.2)= (1.8, 2.2)$. This can be seen in the x-axis
4. In layman’s term, for $\epsilon = 0.6$, there exists a $\delta=0.2$ such that if we take the value a particular  $x$ between $1.8$ and $2.2$, we are sure that the corresponding $f(x)$ is between $10.4$ and $11.6$.  Recall, however,  that $0.6$ is a particular value.  The definition states that we can make it as small as we want and still find a suitable $\delta$, however small, our $\epsilon$ is.
5. In fact a game can be developed where player A gives a particular $\epsilon$, and player B searches for a suitable $\delta$ satisfying the definition. No matter small a value player A assigns to $\epsilon$, we are sure that player B can always find a suitable $\delta$ satisfying the definition.

The general strategy in proving limits by $\epsilon-\delta$ definition is to manipulate the inequality $|f(x) - L| < \epsilon$ such that the expression $|f(x)-L|$ is simplified to $|x - a|$.

To explain the$\epsilon-\delta$ definition of limits further, I will give you three or four more examples in the near future.

15 thoughts on “Limit by epsilon-delta proof: Example 1”

1. You might want to fix up #1 in the interpretation of the definition. Your Limit and epsilon values are reversed.

2. The epsilon-delta limit definition represent a big learning step for most students, and sure is a real challenge. It is quite important to have a visual representation of the formal and not intuitive definition. An animation might help too, to see how the value of delta is influenced by the value of epsilon.

By the way, you have a small mistake in statement #1 above.

Sylvain Bérubé
Sherbrooke, Québec

3. @Nick and Silvain:

Thank you for pointing out the typo error. Error fixed. 🙂

4. Other than that, I love watching your page. I subscribe through RSS feed on Google Reader. This is one of my favorites you’ve posted.

5. Hey thanks for including my GeoGebra links on your page. I’ve been looking for new ideas to put together on GeoGebra as a hobby. Let me know if you have an idea you’d like to see and I’ll try to do it the best I can. Thanks so much.

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• Thank you Funsho. 🙂

8. Can you please state an example showing that the limit does not exist at a point of discontinuity. I mean you can assume a point of discontinuity like a jump(break) or a point where value of function tends to infinity. That would really help.