What is the argument behind: 1/(1+x) can be approximated by 1-x if x&lt;&lt;1 ?  I couldn't find where this approximation comes from/derived.
123424
A good approximation of $(1+x)^n$ is $1+xn$ when $|x|n &lt;&lt; 1$.  Does this approximation have a name?  Any leads on estimating the error of the approximation?
32266
What I am thinking was I need two formulas, AX.p(X) => something AX.p(X) => ~ something   I guess something maybe is the p(x) and the other is ~p(x) since we was given EX.~p(x)..But actually it can't work for me. And I guess maybe is EX.~p(x) and ~EX.~p(x)..But if I am doing that then the it will go to deadend..so I give up..;(  Wish someone can help me.   For note about EE: 1: EX.p(X) 2: AX.(p(X) => something) 3: something EE1,2 But something cann't have the free variable from p..in this sample, then soemthing cannot including X..so if something is p(X) it cannot use EE, but if p(Y) then can use EE.
72656
While I was reading the book Exploratory examples for real analysis, I came across this:     to show that : $$\lim_{x\rightarrow x_o}\frac{1}{h(x)} = \frac{1}{\lim_{x\rightarrow x_o}{h(x)}} = \frac{1}{M}$$      Discussion:Let $\epsilon \gt 0$.we want to find $\delta \gt 0$ such that       if  $0 \lt |x-x_0| \lt \delta $,then $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|\lt \epsilon$.      In order to get expression involving $|h(x)-M|$,we first find the common denominator:        $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|=\Big|\frac{M-h(x)}{M~h(x)}\Big|$      $~=\Big|\frac{h(x)-M}{|M|~|h(x)|}\Big|$.      As we have to deal with variable term $|h(x)|$ in denominator.We can construct bound .Let $\epsilon =\frac{|M|}{2}$ ,since lim$_{x\to x_o}h(x)=M$,there exists $\delta_1\gt 0~~$ s.t.       if $0 \lt |x-x_0| \lt \delta_1 $,then $|h(x)-{M}|\lt \frac{|M|}{2}$.      As you study steps below,identify that step where $h(x)\neq 0$ for all $0 \lt |x-x_0| \lt \delta_1 $ plays a critical role:      $|M|=|M-h(x)+h(x)|$      $~~~~~~\leq |M-h(x)|+|h(x)|$      $~~~~~~= |h(x)-M|+|h(x)|$      $~~~~~~\lt \frac{M}{2}+|h(x)| \implies$      $\frac{|M|}{2} \lt |h(x)| \implies$      $\frac{1}{h(x)}\lt \frac{2}{|M|}$,where $0 \lt |x-x_0| \lt \delta_1 $.      Now that we've bound for $1/h(x)$,we can apply the definition to term $|h(x)-M|$.      Since     lim$_{x\rightarrow x_o}h(x)=M$,we can find $\delta_2 \gt 0$ such that,       $~~~~~~~~~~~~~~~~~~~~~~~~~~~~$if  $0 \lt |x-x_0| \lt \delta_2$, then $|h(x)-M|\lt \frac{|M|^2 \epsilon }{2}$   The thing I can't understand is for what purpose did the author work with such complicated expression involving $\epsilon $ in the end of the discussion?
29195
I have looked up various sites online. They only explain it in a very basic arithmetic way. Would someone explain to me why is $\displaystyle\frac{x}{\frac{1}{2}}$ is $\displaystyle\frac{2x}{1}$?
78540
I am trying to prove that for a continuous $f\colon[1,\infty)\rightarrow\mathbb R$ and $f(x)\to a$ as $x\to\infty$ it could be approximated by $g(x)=p(1/x)$ where $p$ is a polynomial.
32090
Problem: I need to formally prove that $$\lim_{x\to 1} x^3 = 1.$$ My work: This is what I have so far and I'm generally a bit stuck with these proofs from here onwards. Because$$-\epsilon &lt; x^3-1 &lt; \epsilon  =  | x^3-1 | &lt; \epsilon,$$ then $$ -\epsilon &lt; x^3-1 &lt; \epsilon$$ $$-\epsilon+1 &lt; x^3 &lt; \epsilon +1$$ $$ \sqrt[3]{-\epsilon+1}&lt; x &lt; \sqrt[3]{\epsilon+1}.$$ Hoping that what I have done so far is correct. Am I right in thinking that $$ \sqrt[3]{-\epsilon+1}&lt; x &lt; \sqrt[3]{\epsilon+1}$$ is giving me an interval where $x$ is going to give me a $f(x)$ value that falls within the distance $\epsilon$ from the limit on the $y$-axis ? Or is this interval smaller than the $\epsilon$-distance on the $y$-axis?
88650
I was searching for the reason why Stirling's Approximation holds true. I found the website  which apparently shows why this is the case. Does this part of the equation make sense in the proof? If so why?  $$\int_{1}^{n}\ln(x)dx\approx\sum_{k=1}^{n}\ln(k)$$
14454
Claim: If $m/n$ is a good approximation of $\sqrt{2}$ then $(m+2n)/(m+n)$ is better.   My attempt at the proof:   Let d be the distance between $\sqrt{2}$ and some estimate, s.  So we have $d=s-\sqrt{2}$   Define $d'=m/n-\sqrt{2}$ and $d''=(m+2n)/(m+n)-\sqrt{2}$  To prove the claim, show $d''&lt;d'$  Substituting in for d' and d'' yields:  $\sqrt{2}&lt;m/n$   This result doesn't make sense to me, and I was wondering whether there is an other way I could approach the proof or if I am missing something.
30887
I want to prove that: $\ln(x+1)&lt; x$.  My idea is to define:  $f(x) = \ln(x+1) - x$, so:  $f'(x) = \dfrac1{1+x} - 1  = \dfrac{-x}{1+x} &lt; 0, \text{ for }x &gt;0$.  Which leads to $f(x)&lt;f(0)$, so $\ln (x+1)-x&lt;0$.  Is that a valid proof? Any other ideas?  Thanks.
76726
$$\frac{1}{n+1}&lt; \log(1+ 1/n)$$  Any ideas? I tried estimating the difference between $1/n$ and the logarithm and comparing with $1/n-1/(n+1)$ but I miserably failed.
98417
So we want to find a $\delta&gt;0$   such that for all $2-\delta&lt;x&lt;2+\delta$  , we will have $4-\epsilon&lt;x^{2}&lt;4+\epsilon$ for all $\epsilon&gt;0$  . If we can find a way to express $\delta$  as a function $\delta (\epsilon)$,  $\delta:\mathbb{R}_+ \to \mathbb{R}_+ $ then we will have solved the problem. But I can't see how to relate $\epsilon$ to $\delta$ in this case.  My initial reaction is write $(2-\delta)^2&lt;x^2&lt;(2+\delta)^2$, this makes the two inequalities look related, but I am not sure where to go from here.
49038
I know that $\lim_{x\to0}\frac{x}{x}=$ 1.  But in my text book, it is written that it is $\infty$ and even $\frac10=\infty$.  But how is it possible? And are they both same? What is the difference between indeterminate and undefined?  And if they are $\infty$ then what is the reason? And why can't we define something for these numbers?    Now see that book says when $x=0$, $t=\infty$. If it is wrong, why did they just write it?
22913
One of fundamental inequalities on logarithm is: $$ 1 - \frac1x \leq \log x \leq x-1 \quad\text{for all $x &gt; 0$},$$ which you may prefer write in the form of $$ \frac{x}{1+x} \leq \log{(1+x)} \leq x \quad\text{for all $x &gt; -1$}.$$  The upper bound is very intuitive -- it's easy to derive from  as follows: $$ \log(1+x) = \sum_{i=1}^\infty (-1)^{n+1}\frac{x^n}{n} \leq (-1)^{1+1}\frac{x^1}{1} = x.$$  My question is: "what is the intuition behind the lower bound?" I know how to prove the lower bound of $\log (1+x)$ (maybe by checking the derivative of the function $f(x) = \frac{x}{1+x}-\log(1+x)$ and showing it's decreasing) but I'm curious how one can obtain this kind of lower bound. My ultimate goal is to come up with a new lower bound on some logarithm-related function, and I'd like to apply the intuition behind the standard logarithm lower-bound to my setting.
111607
The intuitive definition for $\lim\limits_{x \to a} f(x) = L$ is the value of $f ( x )$ can be made arbitrarily close to $L$ by making $x$ sufficiently close,  but not equal to, $a$ . I can easily understand this ,but for the (ε, δ)-definition of limit:For every real ε > 0, there exists a real δ > 0 such that for all real x, 0 &lt; | x − a | &lt; δ implies | f(x) − L | &lt; ε. Oh, god, I cannot understand it completely.  As it is the formal definition of limit, I think it should be precise but somewhat should include the mean of the above intuitive definition, so as for "$f ( x )$ can be made arbitrarily close to $L$" in the intuitive definition correspond to “For every real ε > 0,| f(x) − L | &lt; ε” in the formal definition, it's fine! but does  “making $x$ sufficiently close,  but not equal to, $a$ .” correspond to “there exists a real δ > 0 such that for all real x, 0 &lt; | x − a | &lt; δ” ? This is the point I cannot understand, because I am not sure if  “there exists a real δ > 0 such that for all real x, 0 &lt; | x − a | &lt; δ”  shows "x close enough, but not equal, to $a$".  The other question is: does the biggest δ also get smaller as ε is getting smaller? Why? (Exclude the case when f(x) is a constant function.)  Why do we need the formal definition of a limit? Does the intuitive definition have some flaw?  P.S. Thank you everyone, but I must declare I only have some basic knowledge of limit, I only started to learn calculus a few days ago.
55949
Currently I read about the  of . Mainly it's the problem that in the equation $X-X$ for $X$ being an interval the following is calculated: $$X-X=\{x-y:x\in X, y\in X\}$$ instead of $\{x-x:x\in X\}$. Both $X$ in $X-X$ are treated as different inputs. Thus $X-X\neq \{0\}$ so that there is no additive Inverse in interval arithmetic.  My Question: Is there a known solution for the dependency problem such that there is an additive (an multiplicative) inverse in interval arithmetic? Is there an extended version of interval arithmetic where addition is a group?  What I have found so far: I read about the centered form. For example, instead of calculating with $f(x)=x^2+x$ one can calculate with $(x+\tfrac 12)^2-\tfrac 14$ where $x$ only occurs once. The centered form helps, but does not provide an additive inverse...
112786
I've started studying calculus.  As part of studies I've encountered a question. How does one prove that $1 &gt; 0$?  I tried proving it by contradiction by saying that $1 &lt; 0$, but I can't seem to contradict this hypothesis.  Any help will be welcomed.
14087
Find $\lim_\limits{x\to 1^-}{\ln x\cdot \ln(1-x)}$. I can't even start because I don't really know what $x\to 1^-$ means. If you know what it means it would really help me. I would as well if you helped me with finding the limit, hinted or thorough.