Use the definition of limit to prove that $\lim_{x \to 0} \frac{x}{x+1}=0$ Use the definition of limit to prove that $\lim_{x \to 0} \dfrac{x}{x+1}=0$
My attempt:
Let $\epsilon >0$ then I need to find $\delta>0$ such that if $0<|x|<\delta$ then $\left| \dfrac{x}{x+1} \right| < \epsilon$
What I was thinking is that, to get a $\delta$ that satisfies the definition, I should see what $\epsilon$ satisfies $|x|<\epsilon |x+1|$ and then I could define $\delta$ in terms of such an $\epsilon$.
If $\epsilon =1$ then: $0\leq|x|<|x+1|$ implies that $x^2 < x^2+2x+1$ and then $x>-\dfrac{1}{2}$. In particular, if $-\dfrac{1}{2}<x<\dfrac{1}{2} \Rightarrow |x|<\dfrac{1}{2} \Rightarrow \dfrac{1}{2} < x+1<\dfrac{3}{2}$ so that $|x+1|=x+1$ and then $\dfrac{1}{2}<|x+1| \Rightarrow \dfrac{1}{2|x+1|}<1$. Since $|x|<\dfrac{1}{2}$, then $\left| \dfrac{x}{x+1}\right| = \dfrac{|x|}{|x+1|} <\dfrac{1}{2|x+1|}<1$.
So $\delta=\dfrac{\epsilon}{2}$ stisfies the definition.
If $\epsilon \neq 1$ then:
$0 \leq |x|<\epsilon |x+1|$ Implies that: $x^2<{\epsilon}^2(x^2+2x+1) \Rightarrow (\epsilon ^2-1)x^2+2 \epsilon ^2x + \epsilon ^2 > 0$.
If $0< \epsilon <1 \Rightarrow \epsilon^2-1<0 \Rightarrow x^2 + \dfrac{2 \epsilon^2}{\epsilon^2-1}x+\dfrac{\epsilon^2}{\epsilon^2-1}<0 \Rightarrow \left(x+ \dfrac{\epsilon}{\epsilon-1} \right) \left(x+ \dfrac{\epsilon}{\epsilon+1} \right) < 0$
Let $ x_1 = -\dfrac{\epsilon}{\epsilon-1}$, $x_2=-\dfrac{\epsilon}{\epsilon+1}$. We see that $x_1 >0>x_2$. Then solving the inequality for $x$ we would have $x_2=-\dfrac{\epsilon}{\epsilon+1} < x < -\dfrac{\epsilon}{\epsilon-1} = x_1$.
I am expecting to get something like: $-\phi(\epsilon) < x < \phi(\epsilon) \Rightarrow |x|< \phi(\epsilon)$ on each case so that $\delta = \phi(\epsilon)$ satisfies the definition, but I don't know how to continue here. Could someone give me a hint? Maybe there is a simpler way to approach the exercise.
 A: Suppose $\delta < \min\{\frac12, \frac{\epsilon}{2}\}$. Then $|x + 1| \ge \min\{\frac12, 1 - \frac{\epsilon}{2}\} \ge \frac12$ so that $\frac{|x|}{|x+1|} \le 2\delta < \epsilon$.
A: Premise
$0 < |x| < \delta \implies -\delta < x < \delta.$
To Do
Establish a relationship between $0 < \epsilon$ and $\delta$ such that the premise will imply that 
$\displaystyle \left|\frac{x}{x+1}\right| < \epsilon \iff -\epsilon < \frac{x}{x+1} < \epsilon.$

In addition to establishing a relationship between $\delta$ and $\epsilon$, I will impose the artificial constraint that 
$\displaystyle \delta \leq (1/2) \implies -\frac{1}{2} \leq x \leq \frac{1}{2} \implies \frac{1}{2} \leq (x + 1) \leq \frac{3}{2}.$
Note that the constraint $~\displaystyle \delta \leq \frac{1}{2}~$ forces $(x+1)$ to always be positive, which simplifies the analysis.

To find the minimum value of $~\displaystyle \frac{x}{x+1}$, since the numerator will be negative, and the denominator will be positive, you will want to minimize both the numerator and denominator.
Therefore $~\displaystyle -\frac{\delta}{(1/2)} < \frac{x}{x+1}.~$
To find the maximum value of $~\displaystyle \frac{x}{x+1}$, since the numerator will be positive, and the denominator will be positive, you will want to maximize the numerator and minimize the denominator.
Therefore $~\displaystyle \frac{\delta}{(1/2)} > \frac{x}{x+1}.~$
From the previous discussion, you have that 
$$- 2\delta < \frac{x}{x+1} < 2\delta. \tag1 $$
Therefore, setting $\displaystyle \delta = \min\left[\frac{1}{2}, ~\frac{\epsilon}{3}\right] \implies $
$$-\epsilon < \frac{x}{x+1} < \epsilon,$$
as required.
A: A proof based on the definition. Let $\epsilon>0$ such that $\epsilon<1$. We are seeking a $\delta$ such that if $|x|<\delta$ we get:
$\dfrac{x^{2}}{x^{2}+2x+1}<\epsilon^{2}$ which is equivalent to,
$\dfrac{\epsilon^{2}-\epsilon}{1-\epsilon^{2}}<x<\dfrac{\epsilon^{2}+\epsilon}{1-\epsilon^{2}}$. Now take $\delta=\dfrac{1}{2}\dfrac{\epsilon-\epsilon^{2}}{1-\epsilon^{2}}$.
If $|x|<\delta$ we get $-\delta<x<\delta$.
Clearly $\delta<\dfrac{\epsilon^{2}+\epsilon}{1-\epsilon^{2}}$. Also
$-\delta>\dfrac{\epsilon^{2}-\epsilon}{1-\epsilon^{2}}$. Thus $|x|<\delta$ implies $\left|\dfrac{x}{x+1} \right|<\epsilon$.
If $\epsilon_{1}\geq\,1$,then clearly for the same $\delta$ we get :
$\left|\dfrac{x}{x+1} \right|<\epsilon$$<1\leq\epsilon_{1}$ and the result is proved!!
A: Choosing a positive number $\delta < \frac{\epsilon}{1 + \epsilon}$ will make $\frac{\delta}{1 - \delta} < \epsilon$; if $0 < |x| < \delta$, then by the reverse triangle inequality $$\left|\frac{x}{x+1}\right| \le \frac{|x|}{1 - |x|} < \frac{\delta}{1 - \delta} < \epsilon$$
