Real Analysis Sequence limit

Consider the following sequence:

$$\sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2+\sqrt{2+\sqrt{2}}} \cdots$$

a) Prove by induction that all terms of the sequence are bounded above by two.

b) Show that this sequence converges.

c) Calculate it's limit.

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I'm having trouble with (c). How can I do it using the definition of a limit of a sequence?

Here are my answers for (a) and (b):

(a)

The sequence can be defined as:
$$x_1 = \sqrt{2}\\ x_{n+1} = \sqrt{2+x_n}$$
Now let's prove by induction. The base case, where $n=1$, we have immediately that $\sqrt{2}<2$.
Let's then suppose that $x_n < 2$. Then we have:

$$x_n + 2 < 2 + 2 = 4\\ x_{n+1} = \sqrt{2+x_n} =\sqrt{x_n+2} < \sqrt{4} = 2$$

(b) Since $x_{n+1} = \sqrt{2 + x_n}$ we can see that $x_{n+1}>x_n$ therefore that's a monotonic sequence. In (a) we showed that it is bounded above, and because $x_{n+1}>x_n$ we can say that $|x_n| < 2$, hence the sequence is bounded above and below. By Monotone Convergence Theorem, the sequence converges.

(c) From definition of limit of a sequence:

$$\forall \epsilon > 0, \exists n_0 \in \mathbb{N} \text{ such that } \forall n > n_0 \rightarrow |x_n - L| < \epsilon$$

Now I need to find L... How can I apply that definition to find the limit?

Answer

In the last point you should not use the $\epsilon$–$\delta$ definition. You know that the limit exists. Let it be $L$. Then

$$\sqrt{2+L}=L$$
or $2+L=L^2$, $L^2-L-2=0$, $(L-2)(L+1)=0$. But $L>0$, so $L=2$.