Suppose we have $x_1=\sqrt{2}$ and we have $x_{n+1}=(\sqrt{2})^{x_n}$. We have to find out the limit of the sequence.  I was observing that $x_2=(\sqrt{2})^\sqrt{2}$   and $x_3=(\sqrt{2})^{{\sqrt{2}}^{\sqrt{2}}}$ and so on now it is clear that the function is increasing. Now how to prove that the sequence is convergent and then how to find the limit? The calculative value is going towards $2$. But how to prove it!!
122604
How can I prove      $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}=2$$   I don't know which method can be used for this?
48433
Hi I have a question regarding of limits to infinity please help! Thank You!  The question states the user to find the following limit:  $  \lim_{n\to\infty} n^2 ({\sqrt[n]{x}-\sqrt[n+1]{x}}) $      Where $(x &gt; 0) $  Thank You!!!
70716
Consider this limit:  $$ \lim_{x \to 1} \frac{\sqrt x -  1}{ \sqrt[3] x -  1}  $$  The answer is given to be 2 in the textbook.  Our math professor skipped this question telling us it is not in our syllabus, but how can it be solved?
53291
Problem  The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation:   $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$.   Evaluate the limit $\lim_{n\to\infty} a_n$.  Solution   Show that the sequence $(a_n)_{n=1}^\infty$ is monotonic. The statement $$V(n): a_n &lt; a_{n+1}$$ holds for $n = 1$, that is $\sqrt2 &lt; \sqrt{2+\sqrt2}$. Let us assume the statement holds for $n$ and show that $V(n) \implies V(n+1)$. We have that $$a_n &lt; a_{n+1}.$$ Adding 2 to both sides and taking square roots, we have that $$\sqrt{2+a_n} &lt; \sqrt{2+a_{n+1}},$$ that is $a_{n+1} &lt; a_{n+2}$ by definition. Find bounds for $a_n$. The statement $$W(n): 0 &lt; a_n &lt; 2$$ holds for $n=1$, that is $0 &lt; \sqrt2 &lt; 2$. Let us assume the statement holds for $n$ and show that $W(n) \implies W(n+1)$. We have that $$0 &lt; a_n &lt; 2.$$ Adding two and taking square roots, we have that $$0 &lt; \sqrt2 &lt; \sqrt{2+a_n} &lt; \sqrt4 = 2.$$ The limit $\lim_{n\to\infty} a_n$ exists, because $(a_n)_{n=1}^\infty$ is a bounded monotonic sequence. Let $A = \lim_{n\to\infty} a_n$. Therefore the limit $\lim_{n \to\infty} a_{n+1}$ exists as well and $\lim_{n \to\infty} a_{n+1} = A$. (For $(n_k)_{k=1}^\infty = (2,3,4, \dots)$, we have that $(a_{n_k})_{k=1}^\infty$ is a subsequence of $(a_n)_{n=1}^\infty$, from which the statement follows.) We have that $a_{n+1} = f(a_n)$. That means that $A = \lim_{n\to\infty} a_n = \lim_{n \to\infty} {f(a_n)} = f(\lim_{n \to\infty} a_n) = f(A) = \sqrt{2 + A}$. Solving the equation $A = \sqrt{2 + A}$, we get $A = -1 \lor A = 2$. Putting it all together, we get that $A = 2$, because the terms of the sequence are increasing and $a_1 &gt; 0$.   Is my solution correct?
87187
I was trying to do this limit $$\lim\limits_{n\rightarrow\infty} \sqrt{n}(\sqrt{n+1}-\sqrt{n})$$  and I cant, I will be pleased if someone give me a hint to do it.
99792
Let $a\ge1$ and define the sequence $(x_n)$ recursively by: $$x_1 = \sqrt{a}$$ $$x_{n+1}= \sqrt{a+x_n}$$  Here's what I did:  Plugging in some values makes it seem as if the sequence is increasing. I first solve the equation $t=\sqrt{a+t}$, equivalently $t^2 -t -a =0$, to get an idea what's going on. The roots are $$t_{1,2}= \frac{1 \pm \sqrt{1+4a} }{2}.$$ Therefore, between the roots, that is for $$t\in \left( \frac{1 - \sqrt{1+4a} }{2}, \frac{1 + \sqrt{1+4a} }{2}\right)$$  we have $t^2-t-a \le 0 \iff t \le \sqrt{a+t}$. Here's my attempt at showing inductively that $$\forall n\in \mathbb{N}, \ x_n \le \frac{1 + \sqrt{1+4a} }{2}$$  For $n=1$, the statement is true (simple arithmetic). Suppose it's true for $n$. Consider  $$x_{n+1}= \sqrt{a+x_n}\le \sqrt{a+\frac{1 + \sqrt{1+4a} }{2}}=\frac{1 + \sqrt{1+4a} }{2},$$  because $\frac{1 + \sqrt{1+4a} }{2}$ satisfies $x=\sqrt{x+a}$.  So this establishes the sequence $(x_n)$ is bounded. Now, by induction again, I try to show it's monotonically increasing. The base case $x_1\le x_2$ is obvious. Assume (strong induction) $x_{n-1}\le x_n$ for all $n&gt;1$. But since we have $$x_1\le x_n \le \frac{1 + \sqrt{1+4a} }{2}, $$   then $x_n$ lies in the appropriate interval, so $x_n \le \sqrt{a+x_n}=x_{n+1}.$  Is this proof correct? Can the condition on $a$ be relaxed to $a&gt;0$? Maybe even small negative $a$? And could someone provide me with another proof of convergence of $x_n$? I absolutely hate this one!
50124
Prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}},  \sqrt{2\sqrt{2\sqrt{2}}} \ $ converges to $2$.   My attempt  I proved that the sequence is increasing and bounded by $2$, can anyone help me show that the sequence converges to $2$? Thanks for your help.
27684
$$\lim_{n\to\infty}\sqrt n(\sqrt[n]x-1)$$ I have never previously encountered functional sequences, so I am a newbie in this matter. I know that intuitively, $x^{1/n}\to1$ and that $\infty\cdot0=1$. As a result, the limit should be 1. However, can this reasoning be considered correct and rigorous enough? I'm pretty sure that if I write this in my first year bachelor exam, my answer will be considered incorrect.   What is the more formal way of doing this? (I think that the squeeze theorem should work, but can't choose suitable bounds). Thanks in advance.
50518
Find all x for which $\sum_{n=1}^n x^{n^2} n!$ is convergent  So I tried using the ratio test and got $\lim_{n\to \infty} x^{2n+1}(n+1)$ but I don't know how to proceed from that.   I also tried using the root test which gave me $\lim_{n\to \infty} x^{n}(n!)^{1/n}$ which wasn't really helpful either...
61480
We define $a_1=1$ and $a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)$.  I want to prove that $a_n$ converges. But first I'd like to show that $a_n&lt;\sqrt{2}$ for every $n\in\mathbb{N}$.  I try to use induction. For a fixed $n$, I suppose that $a_n&lt;\sqrt{2}$.  Any hint to prove that $a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)&lt;\sqrt{2}$?  Thanks.
93101
Find the limit of the sequence {$a_{n}$}, given by$$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{n+1}=\dfrac {1}{3}(1+a_{n}+a^{3}_{n-1}), \ for \ n \ &gt; \ 1$$  My try:  $ a_{1}=0,a_{2}=\dfrac {1}{2},a_{3}=\dfrac {1}{2},a_{4}=0.54$ that is the sequence is incresing and each term is positive. Let the limit of the sequence be $x$. Then $ \lim _{n\rightarrow \infty }a_{n+1}=\lim _{n\rightarrow \infty }a_{n}=x$ $$ \lim _{n\rightarrow \infty }a_{n+1}= \lim _{n\rightarrow \infty }1+a_{n}+a^{3}_{n-1}$$  $\Rightarrow x=\dfrac {1}{3}( 1+x+x^3)$  $\Rightarrow x^3-2x+1=0$  and this equation has three roots $x=\dfrac {-1\pm \sqrt {5}}{2},1$   So the limit of the sequence is $\dfrac {-1 + \sqrt {5}}{2}$.  how can i say that the limit is $\dfrac {-1 + \sqrt {5}}{2}$?
104886
For every $c\in [0;2]$ determine whether the sequence $\{a_n\}_{n\geq 1}$ which is defined as follows:  $a_1=c$, $a_{n+1}=1+\frac{(a_n-1)^2}{17}$ for $n\geq 1$  is monotonic for sufficiently large $n$, and determine whether its limit exits and if it exists, give its value.  I have no idea what to do with this problem. I was able to see that $a_{n+1}-a_n$ is a quadratic function and I also found ot that limit, if exists, is equal to either $1$ or $18$ (that's beacause if $a_n$ id convergent to $g$ then every subsequence is also convergent to $g$). So how to determine the limit for every $c\in [0;2]$?
42804
\begin{align}f(x) &amp;= \dfrac{e^{1/x}}{x^2}\\ x_{n+1} &amp;= f\left( \dfrac{1}{x_n}\right)\end{align} Show that $x_n$ is convergent and that its limit is $0$.    It is very easy to find the limit from the recurrence relation. Just plugging $l$ instead of the terms of the sequence. But how do I show it is convergent?   I tried to evaluate the difference $x_{n+1}-x_n$ but its a messy calculation... Also I tried using the function monotony. But I don't see how it tells me something about the function
64538
I want to use the definition of the limit to show that $\sqrt{n^2+1}-n$ converges to 0.  The definition is as follows: if $\sqrt{n^2+1}-n$ converges to 0, then $\forall \epsilon&gt;0$, there exists an $N&gt;0$ such that $n\ge N \implies \mid\sqrt{n^2+1}-n\mid&lt;\epsilon$.  Now I want to start backwords in order to figure out how to pick N. I know:   $\mid\sqrt{n^2+1}-n\mid=\sqrt{n^2+1}-n$   since $n$ is a natural number and $\sqrt{n^2+1}&gt;n$. So I need to pick an N such that $\sqrt{n^2+1}-n&lt;\epsilon$. I tried multiplying $\sqrt{n^2+1}-n$ by $\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}$ but that didn't really seem to help.   Do you have any ideas on how to find this N?  Thanks
12783
Be $(x_n)_n\ge 1 $  such that $x_1=1$ and $x_{n+1}=x_n+\sqrt{x_n^2+1}$ for every $n\ge 1$. Prove that the sequence $y_n=(2^n/x_n)_n \ge 1$ is convergent and find it s limit.  Being positive and decreasing, $y_n$ is clearly convergent. But finding it's limit really put me intro trouble. Any help?  Thank you!
82145
Prove that $\lim_{n \to \infty} \sqrt[n]a = 1$ using only the basic properties of sequences  I don't have an idea on how to tackle this problem - I thought about the Squeeze Theorem but I can't think of a sequence of n which approaches 1.  Could you give me some hints on how to solve this? If it is possible, please, choose the least sophisticated method to solve this as I am entirely new to Calculus.
104458
Suppose that $$Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} $$and $$P(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}.$$ How do I find $$\lim_{x\rightarrow\infty}\frac{Q(x)}{P(x)}$$ and what does the sequence $$\frac{Q(k)}{P(k)}$$ converge to?  For example, how would I find what the sequence $$\frac{8k^2+2k-100}{3k^2+2k+1}$$ converges to?  Or what is $$\lim_{x\rightarrow\infty}\frac{3x+5}{-2x+9}?$$   This is being asked in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: .