I'm solving a question like this for the first time and I am admittedly kind of confused.
I'm trying to prove that the limit for $f(x) = \sqrt{1+x}$, $x\geq -1$ exists for $x \to 0$.
The limit $L = 1$, which I got by plugging in $x = 0$.
By using the epsilon delta definition of a limit, I get:
$|f(x) - 1|$ < epsilon for $|x - 0|$ < delta $\implies|\sqrt{1+x} - 1|$ < epsilon for $|x|$ < delta.
By manipulating the fraction, I get: $|\sqrt{1+x} - 1| = |(\sqrt{1+x} - 1)(\sqrt{1+x} + 1)/(\sqrt{1+x} + 1)| = |x/(\sqrt{1+x} + 1)|$
Since $x\geq -1$, I get that:
$|x/\sqrt{1+x} + 1| \leq |x|$ < delta
Therefore, if epsilon = delta, then $|\sqrt{1+x} - 1|$ < epsilon for $|x|$ < delta.
My question is, am I on the right track?
Note:
After completing my answer, I took a more careful look at the analysis done by the original poster.
I find his analysis creative, and leading to a more elegant solution than my response, below. I left my approach intact, because it represents what would automatically be my first try.
That is, in a challenging problem like this, I give priority to simplicity, rather than elegance ...to each his own.
Note that the original poster's final specification of
$$\delta = \epsilon$$
must work, when $0 < \epsilon < 1,$ because
my specification was
$$\delta = 2\epsilon - \epsilon^2 = \epsilon \times (2 - \epsilon) > \epsilon.$$
For someone new to these types of problems, this particular problem is no walk in the park.
To Prove: $~\displaystyle \lim_{x \to 0} \sqrt{x+1}$ exists.
First of all, as you noted, because the function $~\displaystyle\sqrt{x+1}~$ is a continuous function, you have that the limit equals $~\displaystyle\sqrt{0+1} = 1.$
The easiest way to prove that the limit exists, via $~\epsilon,\delta~$ is to prove that the limit equals $~1.$
$\underline{\text{Tools}}$
Tool-1:
Given a positive real number $r$, and a variable $x$, you have that:
$|x| < r \iff -r < x < r.$
$|x| > r \iff (x < -r) ~~\text{or}~~ (x > r).$
$|x| = r \iff (x = -r) ~~\text{or}~~ (x = r).$
Tool-2
Given any two non-negative numbers $~r,s~$ you have that $r < s \iff r^2 < s^2.$
This second tool is a consequence of the fact that the function $~f(x) = x^2 ~: x \geq 0~$ is a strictly increasing function. That is, as $~x~$ increases, so does $~f(x).$
To show:
$$\forall ~\epsilon > 0, ~\exists ~\delta > 0 ~\text{such that}~ \left\{ ~|x - 0| < \delta \implies |\sqrt{x+1} - 1| < \epsilon ~\right\}. \tag1 $$
As a consequence of the first tool, the problem may be interpreted as requiring that you find a way to express $~\delta~$ as a function of $~\epsilon~$, so that:
whenever
$$- \delta < x < \delta \tag2 $$
this implies that
$$ - \epsilon < \sqrt{x+1} - 1 < \epsilon. \tag3 $$
The required inequality in (3) above may be re-expressed as the requirement that
$$1 - \epsilon < \sqrt{x+1} < 1 + \epsilon. \tag4 $$
Here, you can assume, without loss of generality, that any positive $\epsilon$ chosen will always be less than $~1.~$ This is because if the inequality in the RHS of (1) holds for any positive $\epsilon_1,~$ then the inequality will also hold for any $~\epsilon > \epsilon_1.$
Therefore, the second tool may be applied to the inequality in (4) above. This allows the requirement to be re-interpreted as
$$(1 - \epsilon)^2 < x+1 < (1 + \epsilon)^2. \tag 5 $$
So, the entire problem has been reduced to finding a relationship between $~\delta~$ and $~\epsilon~$ so that the inequality in (2) causes the inequality in (5) to be true.
Studying the inequalities in (2) and (5),
suppose that
$$(1 - \epsilon)^2 \leq (1 - \delta) ~~\text{and}~~ (1 + \delta) \leq (1 + \epsilon)^2. \tag6 $$
Then, if the inequalities in (6) held, then you would have that
$$(1 - \epsilon)^2 \leq (1 - \delta) < x+1 < (1 + \delta) \leq (1 + \epsilon)^2. \tag7 $$
The inequalities in (7) would imply that whenever (2) holds, then (3) must also hold.
So, the entire problem has been reduced to finding a relationship between $~\epsilon~$ and $~\delta~$ so that the inequalities in (6) are always true.
(6) may be re-expressed as
$$(-2\epsilon + \epsilon^2) \leq -\delta, ~~\delta \leq (2\epsilon + \epsilon^2). \tag 8$$
In order to force (8) to always be true, simply set
$$\delta = 2\epsilon - \epsilon^2 = \epsilon \times (2 - \epsilon). \tag9 $$
Since it has been assumed, without loss of generality, that $0 < \epsilon < 1,$ you have that the expression for $\delta,$ as a function of $\epsilon$ expressed in (9) above, will always be positive.
$\underline{\text{In summary}}$
For any $\displaystyle 0 < \epsilon < 1,~$ set $~\delta = 2\epsilon - \epsilon^2.$
This causes the inequalities in (8) to always hold.
Therefore, the inequalities in (6) always hold.
Therefore, as shown in (7), and as a consequence of the tools, the inequalities in (2) imply the inequalities in (3).
Therefore, the implication in (1) has been established, as required.