This limit question is an eye opener for me.where is the problem in attachment .guide
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This problem comes from our text book.   Evaluate  $$ \lim\limits_{x \to 0}\frac{2^x-1-x}{x^2} $$ without using either L'Hopital's rule or Taylor series.  The picture below shows the solution given by the text book.    Many of the people who responded to this question say that the limit diverges. But according to the solution (see the attached image) given by the text book the limit is  $ \frac{(\ln2)^2}{2} $. Is there an error in the solution given in the attached image?  Note: This problem ( and the solution) comes from Cengage, a long time favourite of the students preparing for IIT JEE  Exam (IITs are India's most prestigious engineering institutes.)  It is suprising that no student/teacher has found this mistake all these days. However, once posted on Stackexchange, the fallacy was resolved in a very short time.  Thanks to all the people who responded to this problem. Now the mistake in the book's solution is found.
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Can you please explain why \begin{align*} \lim_{x\to 0^+}{x^{x^x}} &amp;= 0 \end{align*}
12424
I've got a limit which puzzle me several days. The question is  $$ \lim_{n\to+\infty} \prod_{k=1}^n\left(1+\frac{k}{n^2}\right).$$  Can you help me? Thank you in advance
26412
I have been in a deep confusion for about a month over the topic of limits! According to our book, the limit at $a$ is the value being approached by a function $f(x)$ as $x$ approaches $a$.  I have a doubt that in real number line we can never ever reach the closest value to $a$ because always a more closer value will exist.  Now when talking about our methods for calculating limit, what method comes in our mind when we have to calculate where the value approaches? Let's say we have to find what value $f(x)$ approaches when $x$ moves from $[0,a)$.  So we calculate the value of $f(x)$ at $x=0$, say $4$, and then at $x\to a$.   But the problem is that we don't know the value $x\to a$, so we say the value is $a-h$ where $h\to 0$ and calculate the value of $f(a-h)$, say $5-h$.  Now here is where my doubt starts! In the final step we put the value of $h=0$ say in that it is an infinitesimal quantity.  My doubt is that $h$ was tending to $0$ means that it was never equal to zero  maybe it is infinitesimally small, not a stationary value, not an imaginable value, but we know for sure that it is not  equal to $0$. Maybe it is the point closest to zero, but it is not equal to zero and when we use the result $5-h=5$ we are actually making an error which is tending to zero. Maybe the error is very small, but still, there is some error in that we cannot calculate it but we can see that there is this infinitesimal error present.  That means we don't get the exact limiting value or last value of $f(x)$.  $x$ belongs to $[0,a)$ but a value approximate to infinitesimal? Isn't it right! We get an approximated value?
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Help with Limit: $$\lim_{n \to \infty} \frac{(2 n)! e^n (n)^n}{n! (2 n)^{2 n}}$$  EDIT: Posted a similar question just yesterday but bad formating from my part dropped a term, and lead me to strange places. Honestly, I am just plain stuck, I have been hitting my head against it for $2$ days straight. I know the solution should be $2^{-1/2}$ but...  Help would be appreciated. If anyone has a Wolfram Alpha PRO account I am sure that thing churns out the solution as it is something quite basic I just don't see it and its driving me crazy.
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I don't know how to start evaluate this limit, I cannot use L Hopital's rule. Thank you very much for all responses.  $$\mathop {\lim }\limits_{x \to  + \infty } \left({{\arctan \left({4 \over x}\right)} \over {\left|\arcsin \left( - {3 \over x}\right)\right|}}\right)$$
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I'm working on a practice exam for my masters quals and I am having difficulties with the following limit. According to wolfram alpha, it's value is 1/2. Does anyone know how to find this limit?   $$\lim_{n\to\infty} \left[ n-{n\over e}\left(1+{1\over n}\right)^n \right].$$
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$ \lim_{n\to \infty} ( \lim_{x\to0} (1+\tan^2(x)+\tan^2(2x)+ \cdots + \tan^2(nx)))^{\frac{1}{n^3x^2}} $  The answer should be $ {e}^\frac{1}{3} $  I haven't encountered problems like this before and I'm pretty confused, thank you.  I guess we must use the remarkable limit of $ \frac{\tan(x)}{x} $ when $ x $ approaches $ 0 $ by dividing and then multiplaying with $x$.
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A few days ago I found an interesting limit on the "problems blackboard" of my University: $$\lim_{n\to +\infty}\int_{(0,1)^n}\frac{\sum_{j=1}^n x_j^2}{\sum_{j=1}^n x_j}d\mu = 1.$$ The correct claim, however, is: $$\lim_{n\to +\infty}\int_{(0,1)^n}\frac{\sum_{j=1}^n x_j^2}{\sum_{j=1}^n x_j}d\mu = \frac{2}{3}.$$ In fact, following @Tetis' approach we have: $$I_n = \int_{(0,1)^n}\frac{\sum_{j=1}^n y_j^2}{\sum_{j=1}^{n}y_j}d\mu = \int_{(-1/2,1/2)^n}\frac{\frac{1}{2}+\frac{2}{n}\sum_{j=1}^n x_j^2+\frac{2}{n}\sum_{j=1}^n x_j}{1+\frac{2}{n}\sum_{j=1}^n x_j}d\mu,$$ now setting $x_j=-z_j$ and summing the two integrals $$I_n = \int_{(-1/2,1/2)^n}\frac{\frac{1}{2}+\frac{2}{n}\sum_{j=1}^n x_j^2+\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}{1-\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}d\mu$$ follows, so: $$ I_n-\frac{2}{3}=\int_{(-1/2,1/2)^n} \left(-\frac{1}{2}+\frac{2}{n}\sum_{j=1}^n x_j^2\right)\frac{\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}{1-\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}d\mu &lt; 0,$$ $$ \left| I_n-\frac{2}{3}\right|\leq\int_{\sum x_i^2\leq\frac{n}{4}}\left(\frac{1}{2}-\frac{2}{n}\sum_{j=1}^n x_j^2\right)\frac{\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}{1-\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}d\mu,$$ $$ \frac{2}{3}-I_n\leq \frac{n^{n/2}}{2^{n+1}} \int_{\sum x_i^2\leq 1}\left(1-\sum_{j=1}^{n}x_j^2\right)\frac{x_1^2}{1-x_1^2}d\mu.$$ The last bound, anyway, is too crude, since the RHS is $$ \Theta\left(\left(\sqrt{\frac{e\pi}{2}}\right)^n\frac{\log n}{n^{3/2}}\right).$$  My question now is: what is the asymptotic behaviour of $I_n$?  A second one is: can we prove $I_n\geq\frac{2}{3}-\frac{C}{n}$, for a suitable positive constant $C$, without the Central Limit Theorem?    UPDATE: After a few I came out with a  of my own. The challenge is now to give the first three terms of the asymptotics, and possibly a continued fraction expansion for $I_n$.
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I am trying to do an export of approx 30,000 attachments in our Salesforce Org,  Would anyone have been through a similar exercise and can advise best practice or the best way / tool to do this?  Kind Regards  matt
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Can anyone help me with this limit problem without L'Hopital rule and Taylor series? $$\lim_{x\rightarrow\ -\infty}\left(-2\frac{\arctan(x)}{\pi}\right)^x$$
63450
Suppose $|a|&lt;1$ be a real number. How to solve the following limit? $$\lim\limits_{n\to \infty}(1-a)(1-a^2)\cdots(1-a^n).$$
119118
I've been trying to prove the existence of this limit... trying with a bunch of different curves I'm always getting $0$ (I don't know if I'm using the suggestion correctly) but when I try to make an epsilon-delta proof I get stucked because I don't know how to properly bound the denominator.  $$\lim_{(x,y) \rightarrow (0,2)} \frac{x (y-2)^3}{3x - 5(y-2)^4}$$ (Consider curves close to $3x-5(y-2)^4=0$)  My question is: if indeed this limit does exist -and it's $0$- how can I prove it? As I wrote above I don't know how to work around the denominator. I need something smaller than $|3x-5(y-2)^4|$ ... Or maybe the limit does not exist, and then my question would be what curve can you think of to prove it. Thank you for your time!
103529
How do you get the \delta\rightarrow 0 underneath the line underneath the limit.  This is what I have so far  \underline{\lim\limits}_{\delta \rightarrow 0}
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I think it is zero; $$\lim_{n \rightarrow +\infty } \left(\sqrt[m]{\prod_{i=1}^{m}(n+{a}_{i})}-n\right)$$ we can make that steps: $$\lim_{n \rightarrow +\infty } \left(\sqrt[m]{{n}^{m}\prod_{i=1}^{m}\left(1+\frac{a_i}n\right)}-n\right)$$ and then: $$\lim_{n \rightarrow +\infty } \left(n \cdot \sqrt[m]{\prod_{i=1}^{m}\left(1+\frac{{a}_{i}}{n}\right)}-n\right)$$ each bracket aspires to unit so we have:  $$\lim_{n\rightarrow \infty} (n-n)$$ and..  $$\lim_{n\rightarrow \infty} (0) = 0$$
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Problem Statement:- The number of zeroes at the end of the integer   $$100!-101!+\ldots-109!+110!$$   I am having a bit of a trouble in thinking how do I proceed. A little push in the right direction would be appreciated.   And if you are posting a full solution do use the spoiler tag, as sometimes I cant stop myself from seeing the whole solution and lose the chance of thinking through it by myself with a push in the right direction from you guys.  Also, I dont know whether I am using the right tag feel free to correct it if its wrong.
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Could someone give me a hint? I would like to continue to attempt it. The limit to evaluate that I would like a hint on is:  $$\lim_{(x,y,z)\to(0,0,0)}\frac{xy+yz^2+xz^2}{x^2+y^2+z^4}$$