Basic proofs for epsilon delta limits: $\lim_{x \to a} \frac{x}{1+x}=\frac{a}{1+a}$

Question

Hello I am wondering if my approach and understanding of the epsilon delta definition is correct.

For example, if I wanted to use it to show that $$\lim_{x \to a} \frac{x}{1+x}=\frac{a}{1+a}$$ for $a \in \mathbb{R}, a \neq -1$

Then here are my thoughts;

I know that we want that for all $\epsilon \gt 0$ such that $0 \lt |x-a| \lt \delta$ , then $|f(x)-L| \lt \epsilon$ for a suitable $\delta$.

My attempt:

If $|x-a| \lt a$ then $-a \lt x-a \lt a$ and $0 \lt x \lt 2a$

so $1 \lt x+1 \lt 2a+1$

so $\frac{1}{x+1} \lt 1$

and hence,

$|\frac{x}{1+x}-\frac{a}{1+a}|$=$\frac{|x-a|}{|x+1||a+1|} \lt \frac{|x-a|}{a+1}$

So then we could choose $$\delta=\min\{a,\epsilon|a+1|\}$$

and this would prove that for all $|\frac{x}{x+1}-\frac{a}{a+1}| \lt \epsilon$,

$|x-a| \lt \delta$ ?

How does this seem? Does it make sense? If there are mistakes, what should I consider?

Answer 1

You write

we want that for all $\epsilon \gt 0$ such that $0 \lt |x-a| \lt \delta$ , then $|f(x)-L| \lt \epsilon$ for a suitable $\delta$.

What you actually want is: For every $\epsilon > 0$, there corresponds a $\delta > 0$ such that for all $x$, $0 < |x - a| < \delta$ implies $|f(x) - L| < \epsilon$. Also, since $a$ can be zero or negative, you cannot start with the assumption $|x - a| < a$. The $\delta$ you have chosen can be negative for the same reasons.

Using the fact that

$$\frac{x}{1 + x} = 1 - \frac{1}{1 + x}$$

for all $x\neq -1$, we have

$$\left|\frac{x}{1 + x} - \frac{a}{1 + a}\right| = \left|\frac{1}{1 + a} - \frac{1}{1 + x}\right| = \left|\frac{x - a}{(1 + a)(1 + x)}\right| = \frac{|x - a|}{|1 + a||1 + x|}.$$

for all $x\neq -1$. Now since $a \neq -1$, $|1 + a| > 0$. So we may consider first $0 < |x - a| < \frac{|1 + a|}{2}$. By the triangle inequality,

$$|1 + a| = |(1 + x) - (x - a)| \le |1 + x| + |x - a|,$$

thus

$$|1 + x| \ge |1 + a| - |x - a| > |1 + a| - \frac{|1 + a|}{2} = \frac{|1 + a|}{2}.$$

Hence if $0 < |x - a| < \frac{|1 + a|}{2}$, then

$$\frac{|x - a|}{|1+a||1 + x|} < \dfrac{|x - a|}{|1+a|\cdot \frac{|1+a|}{2}} = \frac{2}{(1 + a)^2}|x - a|,$$

which can be made less than $\epsilon$ by having $|x - a| < \frac{\epsilon(1 + a)^2}{2}$. So choose

$$\delta = \min\left\{\frac{|1+a|}{2},\frac{\epsilon(1+a)^2}{2}\right\}.$$