Setting $\delta$ in $\displaystyle\lim_{x\to1}\dfrac{{x}^{2}}{{x}^{2}+1}$

Question

The book provided a limiting value, but I want a challenging question. I decided to use epsilon-delta, but I can't find an appropriate $\delta$ to $\displaystyle\lim_{x\to1}\dfrac{{x}^{2}}{{x}^{2}+1}$. I followed the process in the book.

(Process.)

$\dfrac{{x}^{2}}{{x}^{2}+1}=1-\dfrac{1}{{x}^{2}+1}$

I assumed $\delta=1$ to find the appropriate $\delta(\epsilon)$.

$0<x<2$

$\therefore1<|{x}^{2}+1|<5\iff\dfrac{1}{5}<\bigg|\dfrac{1}{{x}^{2}+1}\bigg|<1$

At this point, I don't see any solution. What part do I need to elaborate on?

Answer 1

The reason you do not see any solution yet is because you are proving the wrong statement. Before writing an $\epsilon$-$\delta$ proof, you need to make an educated guess about what the limit is actually going to be.

Using elementary calculus, it is actually clear that $\lim_{x→1}\frac{x^2}{x^2+1}$ should equal $\frac{1}{2}$, but notice that in your proof, there is no mention of the number $\frac{1}{2}$. This is the first sign that there is something inherently missing in your proof.

Before we fix this mistake, let us review the $\epsilon$-$\delta$ definition of the limit, just so we can remember what we are actually looking for.

Let $\epsilon > 0$. We are looking for a number $\delta>0$ such that $0<|x-1|<\delta$ implies $\left|\frac{x^2}{x^2+1}- \frac{1}{2}\right|<\epsilon$.

It is important to understand that $\epsilon$ stands for any positive real number. Another problem in your proof is that you have assumed that $\epsilon=1$ (which, I believe, you have called $\delta$), but this is not allowed. We are trying to find a $\delta$ for each $\epsilon$, not just for $\epsilon = 1$ (keeping in mind that $\delta$ can and usually does depend on $\epsilon$).

Let's find our $\delta$ now. We want to bound $\left|\frac{x^2}{x^2+1}- \frac{1}{2}\right|$. With some algebra, it is clear that $$\left|\frac{x^2}{x^2+1}- \frac{1}{2}\right|=\left| \frac{x^2-1}{2(x^2+1)} \right|=\frac{|x^2-1|}{2(x^2+1)}.$$

Since $x^2+1≥1$, it follows that $\frac{1}{x^2+1}≤1$, from which it follows that $$\begin{align} \left|\frac{x^2}{x^2+1}- \frac{1}{2}\right|&= \frac{|x^2-1|}{2(x^2+1)} \\ ≤\frac{1}{2} |x^2-1|.\end{align} $$

It is now easy to bound $\frac{1}{2}|x^2-1|$ (review the proof that $\lim_{x→1}x^2=1$ and try to imitate it). This provides you with the appropriate $\delta$, completing the proof.

Answer 2

I'll write this answer since I think your question was well-intentioned (perhaps you're even a self-studier) and you should get some clarification on what $\epsilon-\delta$ actually does.

The function you posted is continuous at $x=1$ so we have that

$$\lim_{x\to1}\frac{x^2}{x^2+1} = \frac{1^2}{1^2+1} = \frac{1}{2}$$

However this isn't a full proof unless you proved the theorem "continuous implies plug the limit in" yourself. Instead consider this scratch work. So far you have "guessed" that

$$\lim_{x\to1}\frac{x^2}{x^2+1}"="\frac{1}{2}$$

seems like a reasonable choice. To prove it, we would need to use $\epsilon-\delta$, which is a kind of game. The goal of this game is to find $\delta$ as some formula (sometimes a function) of $\epsilon$ and we do this by manipulating the term

$$|f(x)-L| < \epsilon$$

to look like

$$|x-a|<\delta$$

In this case, consider

$$\left|\frac{x^2}{x^2+1}-\frac{1}{2}\right| = \left|\frac{x^2-1}{2(x^2+1)}\right| = \left|\frac{x+1}{2(x^2+1)}\right||x-1|$$

Notice that we have factored out a term of the form $|x-a|$. Now we want to find a max for the term in front

$$\left|\frac{x+1}{2(x^2+1)}\right| \leq \left|\frac{x}{2(x^2+1)}\right| + \left|\frac{1}{2(x^2+1)}\right| \leq \frac{1}{4}+\frac{1}{2} = \frac{3}{4}$$

The first term is from AM-GM inequality, $a^2+b^2 \geq 2|a||b|$. The second term is from $x^2+1\geq 1$. Continuing on, this means that

$$\left|\frac{x+1}{2(x^2+1)}\right||x-1| < |x-1|$$

since $\frac{3}{4} < 1$. What this means is we now have the tools to make a proof with $\epsilon-\delta$, and it follows this script:

For every $\epsilon > 0$, choose $\delta = \epsilon$. This means that

$$\delta = \epsilon > |x-1| > \left|\frac{x+1}{2(x^2+1)}\right||x-1| = \left|\frac{x^2}{x^2+1}-\frac{1}{2}\right|$$

Or in other words, this choice of $\delta$ means

$$|x-1|<\delta \implies \left|\frac{x^2}{x^2+1}-\frac{1}{2}\right| < \epsilon$$

and we have successfully proved that

$$\lim_{x\to1}\frac{x^2}{x^2+1}=\frac{1}{2}$$

Answer 3

I think the book gave you the $L=\frac12$ as well. Then start the process:

$$|f(x)-L|<\epsilon$$ $$\left|\frac{x^2}{x^2+1}-\frac12\right|<\epsilon$$ $$\left|1-\frac{1}{x^2+1}-\frac12\right|<\epsilon$$ $$\left|\frac12-\frac{1}{x^2+1}\right|<\epsilon$$ $$-\epsilon<\frac12-\frac{1}{x^2+1}<\epsilon$$ $$\frac12-\epsilon<\frac{1}{x^2+1}<\frac12+\epsilon$$ We can assume that $\epsilon<\frac12$ hence $$\frac2{1+2\epsilon}<x^2+1<\frac 2{1-2\epsilon}$$ $$\sqrt{\frac{1-2\epsilon}{1+2\epsilon}}<x<\sqrt{\frac{1+2\epsilon}{1-2\epsilon}}$$ We substract the $a=1$ and get $$\small-(1-\sqrt{\frac{1-2\epsilon}{1+2\epsilon}})<x-1<\sqrt{\frac{1+2\epsilon}{1-2\epsilon}}-1$$ Now, $$\delta=\min\left\{1-\sqrt{\frac{1-2\epsilon}{1+2\epsilon}}, \sqrt{\frac{1+2\epsilon}{1-2\epsilon}}-1 \right\}.$$ Process finished.

But you can write the process better. Since I put a condition on epsilon to avoid cases.