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Question:
Prove $\lim\limits_{x\to 1} \frac{x^2}{x^2 + 1} = \frac12$ using $\delta$-$\varepsilon$ method.

Source:
I recently came across this post and I am a beginner in understanding $\delta$-$\varepsilon$ proofs. I searched on Google to learn about proving limits using the definition and carefully reviewed several illustrative examples.

Now, in order to ensure my comprehension, I am presenting my solution below and kindly requesting assistance in identifying any mistakes I may have made.

Proof:

Definition: The limit $\lim\limits_{x\to a} f(x) = L$ if for all $\varepsilon > 0$ there exists $\delta > 0$ such that $|f(x) - L| < \varepsilon $ when $0 < |x - a| < \delta$. i.e. we need to prove that, \begin{align} &\left|\frac{x^2}{x^2 + 1} - \frac12\right| < \varepsilon\ {\text{when}} \ 0 < |x-1| < \delta .\\ {\text{or}}& \ \left|\frac{2x^2 -x^2 - 1 }{2(x^2 + 1)} \right| < \varepsilon\ {\text{when}} \ 0 < |x-1| < \delta .\\ {\text{or}}& \ \left|\frac{x^2 - 1 }{2(x^2 + 1)} \right| < \varepsilon\ {\text{when}} \ 0 < |x-1| < \delta .\\ {\text{or}}& \ \left|\frac{(x+1)(x-1)}{(x^2 + 1)} \right| < 2\varepsilon\ {\text{when}} \ 0 < |x-1| < \delta .\\ {\text{or}} &\ \frac{(x+1)}{(x^2 + 1)}|x-1| < 2\varepsilon\ {\text{when}} \ 0 < |x-1| < \delta .\\ \end{align}

We know that $\dfrac{(x+1)}{(x^2 + 1)} \le 1$.

So we just have to show that, $|x-1| < 2\varepsilon $ when $0 < |x- 1| < \delta$ which is true when $\delta = 2 \varepsilon$.

Thus the limit is proved.

In one of the answer by @NinadMunshi in the above linked post, they chose $\delta = \varepsilon$. Does that mean my work is wrong?

Utkarsh
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2 Answers2

1

Let $\varepsilon > 0$. Suppose $\delta' < 1$, then $\mid x - 1 \mid \ < \ 1 \ \Rightarrow \ $ $0 < x < 2$. Note that

$$ 1 < x + 1 < 3 \ \Rightarrow \ \dfrac{1}{3} < \dfrac{1}{x + 1} < 1 \quad \text{and} \quad \ 0 < x^2 < 4 \ \Rightarrow 2 < 2(x^2 + 1) < 10. $$

$$\dfrac{2}{3} \ < \ \dfrac{2(x^2 + 1)}{(x + 1)} \ < \ 10.$$

$$\Rightarrow \dfrac{2\varepsilon}{3} \ < \ \dfrac{\varepsilon2(x^2 + 1)}{(x + 1)}.$$

Therefore, choose $\delta = min\bigg\{1, \dfrac{2\varepsilon}{3}\bigg\}$. Following that

$$0 \ < \ \mid x - 1 \mid \ < \ \delta \ \Rightarrow \ \mid x - 1 \mid \ \cdot \ \bigg\lvert\dfrac{x + 1}{2(x^2 + 1)}\bigg\lvert \ = \ \bigg\lvert \dfrac{x^2}{x^2 + 1} - \dfrac{1}{2}\bigg\lvert \ < \ \dfrac{2\varepsilon}{3} \ \cdot \ \bigg\lvert\dfrac{x + 1}{2(x^2 + 1)}\bigg\lvert \ < \ \ \dfrac{\varepsilon2(x^2 + 1)}{(x + 1)} \cdot \ \bigg\lvert\dfrac{x + 1}{2(x^2 + 1)}\bigg\lvert \ = \ \varepsilon. $$

Gleberson Antunes
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  • @Utkarsh Yes, you are correct. I made a basic math mistake. I already made the change. Sorry about that. – Gleberson Antunes Aug 12 '23 at 17:50
  • Thanks for that. I'm trying to understand your answer. I'm super beginner to this topic. – Utkarsh Aug 12 '23 at 17:51
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    @Utkarsh I am also a beginner in this matter. That's why I'm trying to help in some way, since I can learn this subject somehow from this. – Gleberson Antunes Aug 12 '23 at 17:53
  • $\delta = min\bigg{1, \dfrac{2\varepsilon}{3}\bigg}$ What is this expression for? (Sorry if it's a dumb question) – Utkarsh Aug 12 '23 at 17:54
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    Initially, we assumed a delta' less than 1. This allowed us to do algebraic manipulations involving x. These manipulations are only valid as long as our delta' is less than 1. So we take this new delta less than 1 and $\dfrac{2\varepsilon}{3}$. – Gleberson Antunes Aug 12 '23 at 17:59
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Your method is ok (except $|\frac{x+1}{x^2+1}|<2$, with $1$ not working), but I think the writing can be simplified as follows:

$$\Bigg|f(x)-f(1)\Bigg|=\Bigg|\frac{x^2}{x^2+1}-\frac 12\Bigg|=\frac 12\Bigg|\frac{x^2-1}{x^2+1}\Bigg|\le \frac 12\underbrace{|x-1|}_{<\ \delta}\underbrace{|x+1|}_{<\ \delta+2}<\epsilon$$

  • we use $x^2+1\ge 1$ so $\frac 1{x^2+1}\le 1$
  • $|x-1|<\delta$ is our hypothesis
  • $|x+1|=|x-1+2|\le|x-1|+|2|<\delta+2$
  • and finally take $\delta=\min(2,\frac\epsilon 2)$ to get $\frac 12\,\delta\,(\delta+2)<\frac 12\times\frac\epsilon 2\times 4=\epsilon$
zwim
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  • I'm not quite able to understand this, I think I need to learn more. I'll be back after some scratch work. – Utkarsh Aug 12 '23 at 17:29
  • I've a doubt. We need to show $\lim_{x\to 1} \frac{x^2}{x^2 + 1} = \frac{1}{2}$. As $x \to 0$, $x^2 \to 0$. So can we transform this limit as $\lim_{x^2 \to 1} \frac{x^2}{x^2 + 1} = \frac{1}{2}$ or $\lim_{y\to 1} \frac{y}{y + 1} = \frac{1}{2}$? Are these type of transformations allowed? And does it make the proof easier in anyway? – Utkarsh Aug 12 '23 at 17:33
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    These transformations are allowed as long as they are continuous because you can compose limits. But most commonly we could do $x\to 1\iff x=1+u$ with $u\to 0$ because this is easier to work with limits in $0$ than in other points. – zwim Aug 12 '23 at 18:34
  • See here for instance https://math.stackexchange.com/a/2486840/399263 – zwim Aug 12 '23 at 18:40