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Functions and single-variable calculus A course resource for Math 218

Ellen Goldstein, Aaron Greicius

Section 1.14 Limits: formal definition

As mentioned before, what makes Definition 1.11.3 less than rigorous is the use of the vague phrases “arbitrarily close” and “sufficiently close”. The “epsilon-delta” formulation given in Definition 1.14.1 is a mathematically precise way of capturing these notions.
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Definition 1.14.1. Limit (formal).

Suppose \(f\) is a function defined everywhere on an open interval containing the point \(a\in \R\text{,}\) except possibly at \(a\) itself. We say that the limit of \(f\) as \(x\) approaches \(a\) exists if there is a value \(L\) satisfying the following property: for all \(\epsilon > 0\text{,}\) there exists a \(\delta > 0\) such that if \(0< \abs{x-a}< \delta\text{,}\) then \(\abs{f(x)-L}< \epsilon\text{.}\) Using logical shorthand:
\begin{equation} \text{for all } \epsilon > 0, \text{ there is } \delta> 0, \text{ s.t., } 0< \abs{x-a}< \delta\implies \abs{f(x)-L}< \epsilon\text{.}\tag{1.74} \end{equation}
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Remark 1.14.2. Anatomy of a definition.

Definition 1.14.1 can come off as forbiddingly technical. This is due to its combination of mathematical and logical details. We endeavor now to unpack and explicate some of these details. We will start with the mathematical nuts and bolts in the interior of the statement and work our way outward through the logical layers.
  • Absolute value.
    The absolute value expressions \(\abs{f(x)-L}\) and \(\abs{x-a}\) are measures of how close \(f(x)\) is to \(L\) and how close \(x\) is to \(a\text{.}\) To say \(\abs{f(x)-L}< \epsilon\) is to say \(f(x)\) is within a distance \(\epsilon\) of \(L\text{;}\) similarly \(\abs{x-a}< \delta\) says \(x\) is within a distance \(\delta\) of \(a\text{.}\)
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  • Conditional.
    The conditional (or if-then) statement
    \begin{equation} 0< \abs{x-a}< \delta\implies \abs{f(x)-L}< \epsilon\tag{1.75} \end{equation}
    asserts that if \(x\) is within a distance \(\delta\) of \(a\) (but not equal to \(a\)), then \(f(x)\) is within a distance \(\epsilon\) of \(L\text{.}\)
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  • Existential quantifier.
    How does adding the existential quantifier “there exists a \(\delta\)” change the meaning? Considering \(\epsilon\) to be fixed for the moment, the statement
    \begin{equation} \text{ there is } \delta> 0, \text{ s.t., } 0< \abs{x-a}< \delta\implies \abs{f(x)-L}< \epsilon\tag{1.76} \end{equation}
    asserts that for all \(x\) sufficiently close (but not equal) to \(a\text{,}\) \(f(x)\) is within a distance \(\epsilon\) of \(L\text{.}\) Indeed, it provides a sort of “safety distance” \(\delta\) and says that as long as \(x\) is within this safety distance of \(a\) (but not equal to \(a\)), then \(f(x)\) is within a distance \(\epsilon\) of \(L\text{.}\)
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  • Universal quantifier.
    Lastly, consider what effect the universal quantifier “for all \(\epsilon\)” has. The full statement
    \begin{equation*} \text{for all } \epsilon > 0, \text{ there is } \delta > 0, \text{ s.t., } 0< \abs{x-a}< \delta\implies \abs{f(x)-L}< \epsilon \end{equation*}
    asserts that for any positive \(\epsilon\text{,}\) the value \(f(x)\) is within a distance \(\epsilon\) for all \(x\) sufficiently close to \(a\text{.}\) In particular, choosing \(\epsilon\) to be as small as you like, we have \(f(x)\) within that small distance \(\epsilon\) of \(L\) for all \(x\) sufficiently close to \(a\text{.}\) In other words, we can make \(f(x)\) arbitrarily close to \(L\) for all \(x\) sufficiently close (but not equal) to \(a\text{.}\)
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Remark 1.14.3. Challenge and response.

It is useful to think of the process of verifying Definition 1.14.1 to establish a limit claim \(\lim_{x\to a}f(x)=L\) as a challenge and response type of procedure.
  • For each positive \(\epsilon\text{,}\) we are challenged to show that for \(x\) sufficiently close to \(a\text{,}\) \(\abs{f(x)-L}< \epsilon\text{.}\) Consider this an epsilon-challenge to the limit statement \(\lim\limits_{x\to a}f(x)=L\text{.}\)
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  • To meet a specific epsilon-challenge \(\epsilon\) we respond by providing a \(\delta\) for which \(0< \abs{x-a}< \delta\) implies \(\abs{f(x)-L}< \epsilon\text{.}\) Consider this proffered \(\delta\) our delta-response that meets the given epsilon-challenge.
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Note that the \(\delta\) we provide in response to a particular epsilon-challenge will typically depend on the particular \(\epsilon\) in question. Indeed, we typically see that the smaller the \(\epsilon\text{,}\) the smaller the \(\delta\text{.}\)
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Let’s see this challenge-response nature of the formal definition of the limit in action. We choose a linear function \(f(x)=ax+b\) for our first example.
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Example 1.14.4. Epsilon-delta: linear function.

Let \(f(x)=-4x+7\text{.}\) Verify that \(\lim\limits_{x\to 1}f(x)=3\) using the epsilon-delta definition of the limit.
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Solution.
We approach the problem in the spirit of the challenge-response characterization given in Remark 1.14.3. Given an arbitrary epsilon-challenge of \(\epsilon > 0\text{,}\) we seek to respond with a suitable \(\delta > 0\) that meets the challenge: i.e., we wish to find a \(\delta\) that guarantees \(\abs{f(x)-3}< \epsilon\) for all \(\abs{x-1}< \delta\text{.}\)
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To do so, we work backwards. Treating \(\epsilon\) as a fixed contstant, we ask what must \(x\) satisfy in order for \(\abs{f(x)-L}< \epsilon\text{.}\) In other words, we should first {\em solve} the inequality \(\abs{f(x)-3}< \epsilon\) for \(x\text{!}\) We have
\begin{align*} \abs{f(x)-3}< \epsilon\amp \iff \abs{-4x+7-3}< \epsilon \\ \amp \iff \abs{-4x+4}< \epsilon\\ \amp \iff \abs{-4(x-1)}< \epsilon\\ \amp \iff \abs{-4}\abs{x-1}< \epsilon \amp \text{(abs. val. prop.)}\\ \amp \iff 4\abs{x-1}< \epsilon\\ \amp \iff \abs{x-1}< \frac{\epsilon}{4} \text{(abs. val. prop.)}\\ \amp \iff -\frac{\epsilon}{4}< x-1< \frac{\epsilon}{4}\\ \amp \iff 1-\frac{\epsilon}{4}< x < 1+\frac{\epsilon}{4}\text{.} \end{align*}
We see then that \(x\) solves the inequality \(\abs{f(x)-1}< \epsilon\) if and only if \(x\in (1-\epsilon/4, 1+\epsilon/4)\text{.}\) But this is true if and only if \(\abs{x-1}< \epsilon/4\text{.}\) Indeed, \(x\in (1-\epsilon/4, 1+\epsilon/4)\) if and only if the distance between \(x\) and \(1\) is at most \(\epsilon/4\text{,}\) if and only if \(\abs{x-1}< \epsilon/4\text{.}\) (See Figure 1.14.6.)
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Summarizing we see that, setting \(\delta=\epsilon/4\text{,}\) we have
\begin{equation*} \abs{f(x)-3}< \epsilon \end{equation*}
for all \(x\) satisfying \(\abs{x-1}< \delta\text{.}\) It follows that for any \(\epsilon\) challenge to the limit statement \(\lim\limits_{x\to 1}f(x)=3\text{,}\) we can find a suitable \(\delta=\epsilon/4\) to satisfy the requirements of the formal definition of the limit. Thus, by Definition 1.14.1, we conclude that \(\lim\limits_{x\to 1}f(x)=3\text{,}\) as desired. for all \(x\) satisfying \(\abs{x-1}< \delta\text{.}\) It follows that for any \(\epsilon\) challenge to the limit statement \(\lim\limits_{x\to 1}f(x)=3\text{,}\) we can find a suitable \(\delta\) meeting this challenge. We conclude that \(\lim\limits_{x\to 1}f(x)=3\text{.}\)
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Remark 1.14.5. Centered intevals.

The final step in the solution to Example 1.14.4 hinged on the observation that \(x\in (1-\epsilon/4,1+\epsilon/4)\) if and only if \(\abs{x-1}< \epsilon/4\text{.}\) The observation is worth generalizing. Given a point \(a\) and distance \(d\text{,}\) call the interval \((a-d,a+d)\) the interval centered at \(a\) with radius \(d\text{.}\) As Figure 1.14.6 illustrates, the interval \((a-d, a+d)\) is precisely the set of all \(x\) whose distance to \(a\) is less than \(d\text{:}\) equivalently, \((a-d,a+d)\) is the set of all \(x\) satisfying \(\abs{x-a}< d\text{.}\)
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Centered interval at a with radius d
Figure 1.14.6. Interval \((a-d,a+d)\) of radius \(d\) centered at \(a\)
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Remark 1.14.7. Epsilon-delta definition.

Note that our polynomial evaluation rule (1.71) could have been applied to the previous example, since the function \(f(x)=-4x+7\) is a polynomial (of degree 1). Indeed, using that rule, we have
\begin{equation*} \lim\limits_{x\to 1}f(x)=f(1)=-4+7=3\text{,} \end{equation*}
in agreement with Example 1.14.4. So what is the purpose of that example? It is important to note that all of the rules in Section 1.12, including the polynomial evaluation rule, were stated without proof. Using the delta-epsilon definition in a particular example gives us a rigorous proof that the result obtained via limit rules is correct.
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Going further, it is possible to give, for each limit rule and formula of Section 1.12, a general proof of its validity using the epsilon-delta definition. Outside of epsilon-delta examples, proofs are not an official topic of this course. For the curious, however, checkout Example 1.14.16 for an example of how we use the epsilon-delta definition to prove limit rules.
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It is possible to give a graphical interpretation of the challenge-response nature of the epsilon-delta definition, and this might help you understand all the logical subtleties of the formal definition pointed out in Remark 1.14.2.
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Example 1.14.8. Visualizing epsilon-delta definition.

The Geogebra interactive below provides a means of visualizing the challenge-response nature of the epsilon-delta proof of a limit claim \(\lim\limits_{x\to a}f(x)=L\) in terms of the graph of \(f\text{.}\) (The window below is a bit narrow. Go to the Geogebra page of the interactive for a larger rendition.) Let’s elaborate on how exactly to parse this graphical representation.
  • The specific \(\epsilon\)-challenge is indicated by a horizontal band centered about \(y=L\text{.}\)
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  • The \(\delta\)-response is indicated by a vertical band centered about \(x=a\text{.}\)
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  • The game, given a specific \(\epsilon\)-challenge, is to find an appropriate \(\delta\) such that when \(x\) is within \(\delta\) of \(a\text{,}\) the values \(f(x)\) are within \(\epsilon\) of \(L\text{.}\) Visually, this is accomplished when all points \((x,f(x))\) on the segment of the graph of \(f\) lying in the vertical band about \(x=a\) also lie within the horizontal band about \(y=L\text{.}\)
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Figure 1.14.9. Visualizing epsilon-delta proofs. Made with Geogebra
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Before moving on to further examples of the epsilon-delta definition it will be worthwhile to fomulate a systematic approach to its application. The steps we used in Example 1.14.4 can be summarized by the more general procedure below.
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Remark 1.14.11. Finding \(\delta\) such that \((a-\delta, a+\delta)\subseteq S\).

In the final step of applying Procedure 1.14.10 to a limit statement \(\lim\limits_{x\to a}f(x)=L\text{,}\) we usually have found an open interval \((c,d)\) containing \(a\text{,}\) and then must find a \(\delta\) such that
\begin{equation*} (a-\delta, a+\delta)\subseteq (c,d)\text{.} \end{equation*}
The following approach is useful in this regard. Observe that
\begin{align*} c \amp =a-(a-c)\\ d \amp =a+(d-a) \end{align*}
and thus \((c,d)=(a-\delta_1, a+\delta_2) \text{,}\) where
\begin{align*} \delta_1 \amp = a-c\\ \delta_2 \amp = d-a\text{,} \end{align*}
are the distances from \(a\) to the left and right endpoints of the interval.
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Visualizing distances to endpoints of interval
Figure 1.14.12. Visualizing distances to endpoints of interval
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Setting \(\delta=\min\{\delta_1,\delta_2\}\) as the minimum of these distances, we now have
\begin{equation*} (a-\delta, a+\delta)\subseteq (a-\delta_1,a+\delta_2)=(c,d)\text{,} \end{equation*}
as desired.
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Depending on the example, we can sometimes easily determine which of \(\delta_1\) and \(\delta_2\) is the minimum, giving us an explicit expression for \(\delta\text{.}\) When things are too complicated, however, it is perfectly fine to set \(\delta=\min\{\delta_1,\delta_2\}\text{.}\)
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Example 1.14.13. Epsilon-delta: radical function.

Let \(f(x)=\sqrt{x-2}\text{.}\) Verify that \(\lim\limits_{x\to 3}f(x)=1\) using the epsilon-delta definition of the limit.
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Solution.
Following (1.77) we treat \(\epsilon\) as an arbitrary positive constant and attempt to solve the inequality
\begin{equation*} \abs{f(x)-1}< \epsilon\text{.} \end{equation*}
We have
\begin{align*} \abs{f(x)-1}< \epsilon\amp \iff -\epsilon < f(x)-1 < \epsilon\\ \amp \iff -\epsilon < \sqrt{x-2}-1 < \epsilon\\ \amp \iff 1-\epsilon < \sqrt{x-2}< \epsilon+1\text{.} \end{align*}
We are tempted now to square all terms in the above inequality to get rid of the radical, but recall that in order for the rule
\begin{equation*} a< b \iff a^2< b^2 \end{equation*}
to apply, we need \(a\) and \(b\) to both be nonnegative. (Example: \(-3< -1\text{,}\) but \(9\not\lt 1\text{.}\)) To deal with this, we will treat two separate cases: \(\epsilon\leq 1\) and \(\epsilon> 1\text{.}\)
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Case: \(\epsilon \leq 1\).

When \(\epsilon \leq 1\text{,}\) all terms in the last inequality above are positive, and we may continue on to conclude
\begin{align*} \abs{f(x)-1}< \epsilon\amp \iff 1-\epsilon < \sqrt{x-2}< \epsilon+1\\ \amp \iff (1-\epsilon)^2 < x-2 < (\epsilon +1)^2\\ \amp \iff 2+(1-\epsilon)^2 < x < 2+(\epsilon +1)^2\\ \amp \iff 3-\epsilon(2-\epsilon)< x < 3+\epsilon(2 +\epsilon)\text{.} \end{align*}
Thus the set of solutions to our original inequality, assuming \(\epsilon\geq 1\text{,}\) is
\begin{equation*} S=(3-\epsilon(2-\epsilon), 3+\epsilon(2 +\epsilon))=(3-\delta_1, 3+\delta_2)\text{,} \end{equation*}
where
\begin{align*} \delta_1 \amp = \epsilon(2-\epsilon)\\ \delta_2 \amp = \epsilon(2 +\epsilon)\text{.} \end{align*}
Note that since \(0< \epsilon\leq 1\text{,}\) both \(\delta_1\) and \(\delta_2\) are positive. It is clear that \(3\in S\text{.}\) Setting \(\delta\) to be the minimum of \(\delta_1, \delta_2\text{,}\) guarantees that \((3-\delta, 3+\delta)\subseteq S\text{.}\) We conclude that for any positive \(\epsilon\leq 1\text{,}\) setting \(\delta=\min\{(\delta_1, \delta_2\}\text{,}\) we have \(\abs{f(x)-1}< \epsilon\) for all \(x\) satisfying \(\abs{x-3}< \delta\text{.}\)
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Note: in fact, it is not difficult to see in this case that
\begin{equation*} \delta_1=\epsilon(2-\epsilon)\leq \delta_2=\epsilon(2+\epsilon) \end{equation*}
and so we could have more explicitly set \(\delta=\epsilon(2-\epsilon)\) in this case. Nonetheless, declaring \(\delta=\min\{\delta_1, \delta_2\}\) works just as well here, as well as in other examples.
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Case: \(\epsilon> 1\).

Now assume \(\epsilon\geq 1\text{.}\) We need to find a \(\delta\) such that \(0< \abs{x-3}< \delta\) implies \(\abs{f(x)-1}< \epsilon\text{.}\) From the previous case (\(\epsilon\leq 1\)), we know there is a \(\delta\) such that \(0< \abs{x-3}< \delta\) implies \(\abs{f(x)-1}< 1\text{.}\) But since \(\epsilon > 1\text{,}\) we have
\begin{align*} 0< \abs{x-3}< \delta \amp\implies \abs{f(x)-1}< 1 \\ \amp \implies \abs{f(x)-1}< \epsilon\text{,} \end{align*}
as desired.
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Our two cases, taken together, show that for all \(\epsilon > 0\text{,}\) there exists a \(\delta> 0\) such that
\begin{equation*} 0 <\abs{x-3}< \delta\implies \abs{f(x)-1}< \epsilon\text{.} \end{equation*}
Thus \(\lim\limits_{x\to 3}f(x)=1\text{.}\)
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As the last example illustrates, as our function \(f\) becomes more complicated, solving the relevant inequality (1.77) can be a delicate and difficult affair. Mindful of this fact, instead of asking you to give a full epsilon-delta proof of a given limit claim, we will sometimes ask you to give a partial proof: namely we will give you one specific \(\epsilon\) as a challenge, and ask you to find a \(\delta\) that satisfies (1.74) for this particular \(\epsilon\text{.}\)
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Example 1.14.14. Finding \(\delta\) for specific \(\epsilon\text{:}\) quadratic function.

Let \(f(x)=1-3x^2\text{.}\) It is a fact that \(\lim\limits_{x\to 1}f(x)=-2\text{.}\) Verify the epsilon-delta definition for this limit statement for the specific epsilon \(\epsilon=\frac{1}{2}\text{.}\)
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Solution.
We wish to find \(\delta\) satisfying
\begin{equation*} 0< \abs{x-1}< \delta \implies \abs{f(x)-(-2)}< \frac{1}{2}\text{.} \end{equation*}
Proceeding as in Procedure 1.14.10, we first solve the inequality \(\abs{f(x)-(-2)}< \frac{1}{2}\text{:}\)
\begin{align*} \abs{f(x)-(-2)}< \frac{1}{2} \amp \iff \abs{(1-3x^2)+2}< \frac{1}{2} \\ \amp \iff \abs{3-3x^2}< \frac{1}{2}\\ \amp \iff \abs{(-3)(x^2-1)}< \frac{1}{2}\\ \amp \iff 3\abs{x^2-1}< \frac{1}{2}\\ \amp \iff \abs{x^2-1}< \frac{1}{6}\\ \amp\iff -\frac{1}{6}< x^2-1< \frac{1}{6}\\ \amp \iff\frac{5}{6}< x^2< \frac{7}{6}\\ \amp \iff\sqrt{\frac{5}{6}}< \abs{x}< \sqrt{\frac{7}{6}} \amp \text{(rad. ineq. prop.)}\\ \amp \iff -\sqrt{\frac{7}{6}}< x < -\sqrt{\frac{5}{6}} \text{ or } \sqrt{\frac{5}{6}}< x < \sqrt{\frac{7}{6}}\text{.} \end{align*}
Notice that in this case, the set of solutions \(S\) to the inequality is a union of two intervals:
\begin{equation*} S=\left(-\sqrt{\frac{7}{6}},-\sqrt{\frac{5}{6}}\right)\cup \left(\sqrt{\frac{5}{6}},\sqrt{\frac{7}{6}}\right)\text{.} \end{equation*}
Notice further that our limit point \(1\) is an element of the second interval, and that we have
\begin{equation*} \left(\sqrt{\frac{5}{6}},\sqrt{\frac{7}{6}}\right)=(1-\delta_1,1+\delta_2) \text{,} \end{equation*}
where
\begin{align*} \delta_1 \amp = 1-\sqrt{\frac{5}{6}} \\ \delta_2 \amp = \sqrt{\frac{7}{6}}-1\text{.} \end{align*}
Thus, setting \(\delta=\min\{\delta_1,\delta_2\}\text{,}\) we have
\begin{equation*} (1-\delta, 1+\delta)\subseteq \left(\sqrt{\frac{5}{6}},\sqrt{\frac{7}{6}}\right)\subseteq S\text{.} \end{equation*}
It follows that
\begin{equation*} 0< \abs{x-1}< \delta \implies \abs{f(x)-(-2)}< \frac{1}{2}\text{,} \end{equation*}
as desired.
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We finish the section with an example illustrating how to use the epsilon-delta definition of the limit to give a rigorous proof of limit rules. As mentioned above, proving limit rules is not an official topic of this course. However, the argument given in the solution below might help further illuminate how the epsilon-delta definition works. We will make use of the famous triangle inequality for the absolute value, which we make official here.
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Proof.

There are a number of different ways of proving the triangle inequality. We will give a proof making use of the identity \(\abs{x}=\sqrt{x^2}\text{.}\) First observe that since both sides of (1.78) are nonnegative, we have
\begin{equation*} \abs{x+y}\leq \abs{x}+\abs{y}\iff (\abs{x+y})^2\leq (\abs{x}+\abs{y})^2\text{,} \end{equation*}
using Theorem 1.8.15. Now begin with \((\abs{x}+\abs{y})^2\text{.}\) We have
\begin{align*} (\abs{x}+\abs{y})^2 \amp = \abs{x}^2+\abs{y}^2+2\abs{x}\abs{y} \\ \amp =(\sqrt{x^2})^2+(\sqrt{y^2})^2+2\abs{x}\abs{y} \\ \amp = x^2+y^2+2\abs{xy} \\ \amp \geq x^2+y^2+2xy \amp (\abs{a}\geq a)\\ \amp = (x+y)^2\\ \amp =(\sqrt{(x+y)^2})^2\\ \amp = \abs{x+y}^2\text{.} \end{align*}
We conclude that \(\abs{x+y}^2\leq (\abs{x}+\abs{y})^2\text{,}\) and thus that \(\abs{x+y}\leq abs{x}+\abs{y}\text{,}\) as desired. Where does the condition for inequality in (1.78) come from? Notice that the chain of equalities above is a chain of equalities if and only if the middle inequality, which uses \(2\abs{xy}\geq 2xy\) is an equality. This in turn is true if and only if \(xy\geq 0\text{,}\) which is equivalent to saying that \(x\) and \(y\) do not have different signs.
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Example 1.14.16. Proof of the sum rule for limits.

Use the epsilon-delta definition of the limit to prove the Sum rule. That is, show that if \(\lim\limits_{x\to a}f(x)\) and \(\lim\limits_{x\to a}g(x)\) exist, then \(\lim\limits_{x\to a}f(x)+g(x)\) exists and satisfies
\begin{equation*} \lim\limits_{x\to a}f(x)+g(x)=\lim\limits_{x\to a}f(x)+\lim\limits_{x\to a}g(x)\text{.} \end{equation*}
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Solution.
Let \(\lim\limits_{x\to a}f(x)=L\) and \(\lim\limits_{x\to a}g(x)=M\text{.}\) We use the epsilon-delta definition to show that \(\lim\limits_{x\to a}f(x)+g(x)=L+M\text{.}\)
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Given an arbitrary \(\epsilon > 0\text{,}\) we must find a \(\delta > 0\) such that
\begin{equation*} \abs{f(x)+g(x)-(L+M)}< \epsilon \end{equation*}
for all \(x\) satisfying \(\abs{x-a}< \delta\text{.}\) Since
\begin{align*} \lim\limits_{x\to a}f(x) \amp = L \amp \lim\limits_{x\to a}g(x)=M\text{,} \end{align*}
we can find positive numbers \(\delta_1\) and \(\delta_2\) such that
\begin{align} \abs{f(x)-L} \amp < \epsilon/2 \amp \abs{g(x)-M}< \epsilon/2\tag{1.79} \end{align}
for all \(x\) satisfying \(\abs{x-a}< \delta_1\) and \(\abs{x-a}< \delta_2\text{.}\) What’s going on here? The \(\delta_1\) and \(\delta_2\) are our responses to the the epsilon challenge of \(\epsilon/2\) to the statements \(\lim\limits_{x\to a}f(x)=L\) and \(\lim\limits_{x\to a}g(x)=M\text{.}\) You will see below, why we chose an epsilon challenge of \(\epsilon/2\) here.
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Letting \(\delta=\min(\delta_1,\delta_2)\) it follows that both inequalities in (1.79) hold for all \(x\) satisfying \(\abs{x-a}< \delta\text{.}\) We claim this \(\delta\) is the response we want to the given \(\epsilon\) challenge to the limit statement \(\lim\limits_{x\to a}f(x)+g(x)=L+M\text{.}\) Indeed for all \(x\) satisfying \(\abs{x-a}< \delta\text{,}\) we have
\begin{align*} \abs{f(x)+g(x)-(L+M)} \amp =\abs{(f(x)-L)+(g(x)-M)}\\ \amp \leq \abs{f(x)-L} +\abs{g(x)-M} \amp (\text{triangle ineq.})\\ \amp < \epsilon/2 +\epsilon/2 \amp (\abs{x-a}< \delta)\\ \amp = \epsilon\text{.} \end{align*}
This proves that given any \(\epsilon > 0\) there is a \(\delta> 0\) such that
\begin{equation*} \abs{f(x)+g(x)-(L+M)}< \epsilon \end{equation*}
for all \(x\) satisfying \(\abs{x-a}< \delta\text{,}\) and hence that
\begin{equation*} \lim\limits_{x\to a}f(x)+g(x)=\lim\limits_{x\to a}f(x)+\lim\limits_{x\to a}g(x)\text{.} \end{equation*}
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