Proving $\lim_{x\to 1} \frac{x^2}{x^2 + 1} = \frac12$ using $\delta$-$\varepsilon$ method

Question

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.

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

Answer 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

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

Answer 2

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 $\frac 12\,\delta\,(\delta+2)<\frac 12\times\frac\epsilon 2\times 4=\epsilon$