# 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.

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.