# Proving Limits

In continuing to use our proof techniques, we will take some time to look more closely at ideas that we have seen before. In particular, we will look at what a limit is formally defined to be. Based on this definition, we will then prove that a limit is indeed what we claim it to be.

## What is a limit?

In Calculus, we use limits extensively. The concept of a limit is used to determine what happens to function values as $$x$$ values get arbitrarily close to, but not equal to, some given point. Without the limit, determining derivatives and integrals would be impossible (well Newton and Leibniz did Calculus without a formal definition of limit, but the work was never finished until the precise definition was given).

As an overview, the limit of $$f(x)$$ as $$x$$ approaches $$c$$ is $$L$$, denoted $$\lim\limits_{x \to c}f(x)=L$$, if, as the $$x$$ values get arbitrarily close to $$c$$, the $$y$$ values get arbitrarily close to $$L$$. As a way to try to make that more clear, the goal is that you would give me a positive number, say $$\epsilon$$ . Based on this number, I would draw a line that far above and that below $$L$$. My goal would then be to find a positive number $$\delta$$. I would then draw a line $$\delta$$ to the right of $$c$$ and $$\delta$$ to the left of $$c$$. This $$\delta$$ would satisfy my needs if $$f(x)$$ stays within the two drawn horizontal lines for all $$x$$ within the vertical lines (see picture below).

Now, instead of just having to do this for one such $$\epsilon$$, we have to be able to do this for any $$\epsilon$$ you could give. The formal definition would then be,
\begin{align*}
\lim_{x \to c}f(x)=L
\end{align*}
if, for any $$\epsilon >0$$, there exists a $$\delta >0$$ such that $$|f(x)-L| < \epsilon$$ for all $$x$$ satisfying $$0 < |x-c| < \delta$$.

## Our limit

Based on this definition, we want to be able to proof that
\begin{align*}
\lim_{x \to 2}(3x-5)=1.
\end{align*}

Instead of jumping right into the proof, we want to look at what $$\delta$$ should be based on a given $$\epsilon$$. Therefore, we start off by simplifying $$|f(x)-L$$.
\begin{align*}
|f(x)-L|&=|3x-5-1| \\
&=|3x-6| \\
&=|3(x-2)| \\
&=|3||x-2| \\
&=3|x-2|.
\end{align*}

Now that we know that $$|f(x)-L|=3|x-2|$$, the point is that, given $$\epsilon$$, we want $$3|x-2|< \epsilon$$ for $$0 < |x-2| < \delta$$. Therefore, we would want to pick $$\delta$$ so that \begin{align*} 3 |x-2| < 3 \delta = \epsilon, \end{align*} in order to ensure that we stay within $$\epsilon$$ of $$L$$. Solving this for $$\delta$$, we see that we should let $$\delta =\frac{\epsilon}{3}$$. Now that we have found our $$\delta$$, we are ready to give our proof.

### Proof

Let $$\epsilon >0$$ be given. Then choose $$\delta =\frac{\epsilon}{3}$$ and $$0 < |x-2| < \delta$$. We then have that \begin{align*} |f(x)-L|&=|3x-5-1| \\ &=|3x-6| \\ &=3|x-2| \\ &< 3 \delta \\ &=3 \frac{\epsilon}{3} \\ &=\epsilon. \end{align*} Hence, $$|f(x)-L| < \epsilon$$ for all $$x$$ satisfying $$0 < |x-2| < \delta$$, so \begin{align*} \lim_{x \to 2}3x-5&=1. \end{align*}

## Conclusion

In this situation we looked more closely at a topic that, by the time you finish Calculus, you should be very familiar with. Instead of just leaving things alone, we took to the time to formally explain the ideas that we had, at many points, taken for granted. In doing so, we were able to more fully grasp a concept and give a proof that what we thought to be true actually was. If you would like more help with proofs, make sure to continue working on the other problems we have posted solutions for in Study Help.

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