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501

The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$. The area of the colored wedge is $\frac{1}{2}x$, and the area of $\triangle ABD$ is $\frac{1}{2}\tan(x)$. By inclusion, we get $$\frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1}$$ Dividing $(1)$ by $\frac{1}{2}\sin(x)$ and taking reciprocals, we get $$\cos(x)\le\frac{\sin(x)}{x}\le1\tag{2}... 431 From a layman's perspective, imagine that I have an infinite number of hotel rooms, each numbered 1, 2, 3, 4, ... Then I give you all of them. I would have none left, so \infty - \infty = 0 On the other hand, if I give you all of the odd-numbered ones, then I still have an infinite number left. So \infty - \infty = \infty. Now suppose that I give you ... 174$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=7^\frac{1}{2}\cdot7^\frac{1}{4}\cdot 7^\frac{1}{8}\cdots=7^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots}=7^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=7$$164 The first sine is in I_1=[-1,1] hence the nth term of the sequence is in the interval I_n defined recursively by I_1=[-1,1] and I_{n+1}=\sin(I_n). One sees that I_n=[-x_n,x_n] where x_1=1 and x_{n+1}=\sin(x_n). The sine function is such that 0\le\sin(x)\le x for every nonnegative x hence (x_n) is nonincreasing and bounded below by zero ... 154 Rewriting the sum as$$ \sum_{n=k+1}^{2k}\frac1n=\sum_{n=k+1}^{2k}\frac1k\cdot\frac1{n/k} $$allows us to identify this as a Riemann sum related to the definite integral$$\int_1^2\frac1x\,dx=\ln 2.$$To see that, divide the interval [1,2] to k equal length subintervals, and evaluate the function f(x)=1/x at the right end of each subinterval. When ... 139 Your only error -- and it's a common one -- is in a subtle misreading of L'Hopital's rule. What the rules says is IF the limit of f' over g' exists then the limit of f over g also exists and the two limits are the same. It doesn't say anything if the limit of f' over g' doesn't exist. 132 The probabilistic way: This is P[N_n\leqslant n] where N_n is a random variable with Poisson distribution of parameter n. Hence each N_n is distributed like X_1+\cdots+X_n where the random variables (X_k) are independent and identically distributed with Poisson distribution of parameter 1. By the central limit theorem, Y_n=\frac1{\sqrt{n}}(... 130 First, and by definition, when dealing with$$\lim_{x\to x_0}f(x)we must assume \,f\, is defined in some neighborhood of \,x_0\, except , perhaps, on \,x_0\, itself, and from here that in the process of taking the limit we have the right and the duty to assume \,x\, approaches \,x_0\, in any possible way but it is never equal to it. Thus, and ... 129 Denote the given problem as x, then \begin{align} x&=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{\cdots}}}}}\\ &=10\cdot10^{\large\frac{1}{2}}\cdot10^{\large\frac{1}{4}}\cdot10^{\large\frac{1}{8}}\cdot10^{\large\frac{1}{16}}\cdots\\ &=10^{\large1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots}\\ &=10^{\large y} \end{align} where y is ... 128 Edited. I justified the application of the dominated convergence theorem. By a simple calculation, \begin{align*} e^{-n}\sum_{k=0}^{n} \frac{n^k}{k!} &= \frac{e^{-n}}{n!} \sum_{k=0}^{n}\binom{n}{k} n^k (n-k)! \\ (1) \cdots \quad &= \frac{e^{-n}}{n!} \sum_{k=0}^{n}\binom{n}{k} n^k \int_{0}^{\infty} t^{n-k}e^{-t} \, dt\\ &= \frac{e^{-n}}{n!} \...

126

You should first prove that for $x > 0$ small that $\sin x < x < \tan x$. Then, dividing by $x$ you get $${ \sin x \over x} < 1$$ and rearranging $1 < {\tan x \over x} = {\sin x \over x \cos x }$ $$\cos x < {\sin x \over x}.$$ Taking $x \rightarrow 0^+$ you apply the squeeze theorem. For $x < 0$ and small use that $\sin(-x) = -\sin x$...

123

A very easy counterexample would be $$1, \underbrace{\frac12, \frac12}_{2\text{ halves}}, \underbrace{\frac13, \frac13, \frac13}_{3\text{ thirds}}, \underbrace{\frac14, \frac14, \frac14, \frac14}_{4\text{ fourths}}, \underbrace{\frac15, \frac15, \frac15, \frac15, \frac15}_{5\text{ fifths}}, \ldots$$ This sequence clearly converges to $0$, but if you try to ...

121

This image of mine seems apropos: In the case of $\lim_{x\to 5} \frac{x^2-25}{x-5}$, the message here is: Away from $x=5$, the function $\frac{x^2-25}{x-5}$ is completely identical to $x+5$; thus, what we expect to find as we approach $x=5$ is the value $5+5$. This anticipated value is what a limit computes. The fact that the original function isn't ...

105

Vadim's answer handles the math (and I've upvoted it), so I will try to provide intuition. The idea is the word "closer" is relative. That is, in some sense, $100{,}000$ is closer to $100{,}010$ than $1$ is to $0$. Of course, in an absolute sense, $1$ is $1$ away from $0$ while $100{,}000$ is $10$ away from $100{,}010$. But, say you had $\$1$and I took ... 96 Update 2/19/2018: It appears that this answer has received a lot of attention, which I'm very glad to know about. When you're reading through this answer and you're trying to learn about$\delta$-$\epsilon$proofs for the first time, I would recommend skipping the sections labeled Addendum. on your first read. Please let me know of any other clarifications ... 95$\newcommand{\QQ}{\mathbb{Q}}$Derivatives don't really go wrong, it's antiderivatives. (EDIT: Actually, the more I think about it, this is just a symptom. The underlying cause is that continuity on the rationals is a much weaker notion than continuity on the reals.) Consider the function$f : \QQ \to \QQ$given by $$f(x) = \begin{cases} 0 & x < \pi \... 94 n! \geq (n/2)^{n/2} because half of the factors are at least n/2. Take n-th root. 85 This is a slightly softer answer. You can 'do calculus' in-so-far as you can define the derivative and perhaps compute some things. But you'll get no theorems out: the main interval theorems (the Intermediate Value Theorem and the Extreme Value Theorem) rely heavily on the fact that the real numbers are complete, which the rationals aren't. In fact, the '... 83 Let$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=x $$Clearly, x>0$$\implies x^2=7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=7x$$Now left is the proof of converge(as conversed with Abdulh Khazzak Gustav ElFakiri) Observe that the rth term T_r of this infinite product is \displaystyle7^{\left(\frac1{2^r}\right)} using Convergence/... 79 Look at this link: http://fatosmatematicos.blogspot.com/2010/08/provas-sem-palavras-parte-20.html Here is the picture I copied from that blog: 79 By considering Taylor series, \displaystyle e^x \geq \frac{x^n}{n!} for all x\geq 0, and n\in \mathbb{N}. In particular, for x=n this yields$$ n! \geq \left( \frac{n}{e} \right)^n .$$Thus$$\sqrt[n]{n!} \geq \frac{n}{e} \to \infty.$$74 A proof I found a while ago entirely relies on creative telescoping. Since \frac{1}{n^2}-\frac{1}{n(n+1)}=\frac{1}{n^2(n+1)},$$\begin{eqnarray*} \sum_{n\geq m}\frac{1}{n^2}&=&\sum_{n\geq m}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\sum_{n\geq m}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\\&+&\frac{1}{6}\sum_{n\geq m}\left(\frac{... 74 Intuition should say: the denominator grows with$n^2$, the numerator grows with$n$. However, the number of fractions also grows by$n$, so the total growth of the numerator is about$n^2$. And that's where intuition stops. From here on, you go with logic and rigor, not intuition. And it gets you to $$\frac1{n^2} + \frac2{n^2}+\cdots + \frac{n}{n^2} = \... 73 You can use \text{AM} \ge \text{GM}.$$\frac{1 + 1 + \dots + 1 + \sqrt{n} + \sqrt{n}}{n} \ge n^{1/n} \ge 1 1 - \frac{2}{n} + \frac{2}{\sqrt{n}} \ge n^{1/n} \ge 1$$73$$2x_n = x_{n-1} + x_{n-2}2x_2 = x_{1} + x_{0}\\ 2x_3 = x_{2} + x_{1}\\ 2x_4 = x_{3} + x_{2}\\ 2x_5 = x_{4} + x_{3}\\ ...\\ 2x_n = x_{n-1} + x_{n-2}$$Now sum every equation and get$$2x_n+x_{n-1}=2x_1+x_0$$Supposing that x_n has a limit L then making n\to \infty we get:$$2L+L=2x_1+x_0\to L=\frac{2x_1+x_0}{3}$$71 First let me tell you that the idea that an infinite sequence "ends with something" is a solid idea. It's a perfectly natural one. The point is that the sequence is not indexed by \Bbb N, anymore, but rather by \Bbb N\cup\{\infty\}, where \infty is another point, which lies after all the natural numbers. The point is that an "infinite sequence" is a ... 67 The limit is not defined because in order for the limit to exist, the value of the function for every possible path to (0,0) must tend to the same finite value. When y = x^2, you have not necessarily shown that the limit is in fact 0. When you transformed to polar coordinates and then took the limit as r \to 0, you are assuming that \theta is a ... 66 Another more general approach:$$a_n\xrightarrow[n\to\infty]{} a\implies \frac{a_1+...+a_n}n\xrightarrow[n\to\infty]{}a$$And since$$\lim_{n\to\infty}\,n\,\sin\frac1n=\lim_{n\to\infty}\frac{\sin\frac1n}{\frac1n}=1\;\;\ldots\ldots$\$

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