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2

$$\sum_{k=1}^n\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)=\frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right)+\mathcal{O}\left(\frac{1}{n^2}\sum_{k=1}^n f\left(\frac{k}{n}\right)^2\right)$$ Since $\lim\limits_{n\rightarrow +\infty}\frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right)^2=\int_0^1f(x)^2dx$, then $\lim\limits_{n\rightarrow +\infty}\frac{... 2$L_1\lim_{x\to 0} 3x \sin \frac{1}{x}=0$by squeez law/ sandwhich theorem as$-x\le x\sin(1/x)\le x.$But the limit$L_2=\lim_{x\to 0}\cos\frac{1}{x}$does not exists because of two accumulation points. Let$x_n=\frac{1}{2n\pi}$and$x'_n=\frac{1}{(n+1/2)\pi}.$Check that$f(x_n)=1$but$f(x'_n)=0$, these two being unequal the limit$L_2$does not exist. ... 0 If the terms of a series do not approach$0$, then the series diverges. If the terms of a series do approach$0$, then the series may or may not converge. The harmonic series is an example of a series that does not converge though the terms approach$0$. See this Wikipedia article. 2 As you have already shown, the eigenvalues of$P$are$\lambda_1 = -1/\sqrt{2}$,$\lambda_2 = 0$,$\lambda_3 = 1/\sqrt{2}$and$\lambda_4 =1$. Therefore, we can write$P$as$P = SDS^T$where$D = \text{diag}(\lambda_1, \lambda_2, \lambda_3, \lambda_4)$. We also know that$P^r = SD^r S^T. Therefore, we have, \begin{align*} \lim_{n \to \infty} M^{(n)} & =... 0 There are several issues with your question. First, you should requireb \gt 0$and$c \ge 0$for it to work properly (e.g., if$c \lt 0$, then$\sqrt{c}$is not even a real number). Second, your$\lt$should be$\le$, e.g., if$b = c = 1$. Third, what you want to prove for "all integers$n$" should be for positive integers$n$since your relation ... 0 Another good way to think is : Using Logarithm Identity -$ \ \ \displaystyle e^{\ln a} = a. \ \ $Thus, $$\displaystyle \left ( \frac {1}{e} (1+x)^{1/x} \right )^{1/x^2} = \ \ e^{\displaystyle \frac {\ln (1+x)-x }{x^3}} .$$ Using Taylor Expansion,$ \ \ \displaystyle \ln (1+x) = x - \frac {x^2}{2} + \frac {x^3}{3} - \frac {x^4}{4} + ... $Thus expression ... 1 HINT: Note that $$4n-1=(2n+1)+(2n-2)=(2n+1)+2(n-1)\,,$$ so there are$n$fractions in the expression. The largest is$\frac1{2n+1}$, and the smallest is$\frac1{4n-1}$, so $$\frac14<\frac{n}{4n-1}\le\frac1{2n+1}+\ldots+\frac1{4n-1}\le\frac{n}{2n+1}<\frac12\,.$$ Now multiply by$n$and think about what happens as$n$increases. 9 Your first approach uses the reasoning$(1+\frac1x)^x$is approximately$e$, so$(1+\frac1x)^{x^2}=[(1+\frac1x)^x]^x$behaves like$e^x$. This is illegal because you're letting$x\to\infty$in your inner expression while preventing$x\to\infty$in the outer expression. There's no rule that allows you to do that -- you can't hold$x$fixed in one part of ... 3 The short answer is that$\lim_{x \to \infty} g(x, f(x)) \ne \lim_{x \to \infty}g (x, \lim_{x \to \infty} f(x))$. (Here,$g(x,y) = \frac{y^x}{e^x}$and$f(x)=(1+1/x)^x$.) See grand_chat's answer for a very stark example of why taking limits of parts of the function before taking the limit as a whole is "illegal." The long answer is that although$(...

1

hint You get a false result because the exponent is $x^2$ , not $x$. $$(1+\frac 1x)^{x^2}=e^{x^2\ln(1+\frac 1x)}$$ $$=e^{x^2\Bigl(\frac 1x -\frac{1}{2x^2}(1+\epsilon(x))\Bigr)}$$ with $$\lim_{x\to+\infty}\epsilon(x)=0$$

0

Hypothesis 1 only says that the functions need to be differentiable near $a$, not differentiable at $a$. Using symbols, they need only be differentiable on some deleted neighbourhood $(a - \delta, a) \cup (a, a + \delta)$ for some $\delta > 0$. This is similar to what we think of for limits: $\lim_{x \to a} F(x)$ only depends on the behaviour of $F(x)$ as ...

2

The Lebesgue-Stieltjes measure associated to $F(x)=\lfloor x\rfloor$ on $(0,\infty)$ is $$\mu(dx)=\sum^\infty_{n=1}\delta_n(dx)$$ where $\delta_k(A)=\mathbb{1}_A(k)$ (total mass $1$ at $k$) So, for any function $f:(0,\infty)\rightarrow\mathbb{R}$ $$\int f\,d\mu=\int f(x)\,\sum^\infty_{m=1}\delta_m(dx)= \sum^\infty_{m=1}\int f(x)\delta_m(dx)=\sum^\infty_{m=... 3 Taking @zwim's hint to work with Riemann-Stieltjes integrals,$$\begin{align}\int_k^{k+1}xd\lfloor x\rfloor&=[x\lfloor x\rfloor]_k^{k+1}-\int_k^{k+1}\lfloor x\rfloor dx\\&=(k+1)^2-k^2-\int_k^{k+1}ndx\\&=k+1\end{align}$$for any integer k\ge0, so$$\int_0^nxd\lfloor x\rfloor=\sum_{k=0}^{n-1}\int_k^{k+1}(k+1)=\tfrac12n(n+1)$$for any integer n\ge0. ... 2 I don't see anything wrong here. If you're worried about some purported fagility of calculus, this doesn't break calculus because you are considering the limits as x\to0^+ of three different functions; in order, of: f:(0,\infty)\to \Bbb R, f(x)=x^0; g:(0,\infty)\to\Bbb R, g(x)=0^x; h:(0,\infty)\to \Bbb R, h(x)=x^x. 1 The answer to this lies in the function f(x,y)=x^y simply put, this function is not continuous at (0,0), and in fact, you have just proven this! What you stated only seems to break calculus because we feel like any function involving basic arithmetic operations should be continuous everywhere, therefore taking limits should decide the values these ... 0 |f(x)|\leq |x|^3 implies that lim_{x\rightarrow 0}f(x)=0 and f is continuous at 0 1 Your proof is perfectly fine just that you don't need \frac{-2}{n}. You can just replace that by 0. Here's another alternative method by using the ratio test: \frac{x_{n+1}}{x_n}=\frac{\frac{2^{n+1}}{n+1}}{\frac{2^n}{n}}=\frac{1}{2}\left(\frac{n+1}{n}\right). Now take the limit on both sides and thus, using the theorem of ratio test for sequences, if ... 1$$\frac n{2^n}=\frac12\frac2{1\cdot2}\frac3{2\cdot2}\frac4{3\cdot2}\frac5{4\cdot2}\cdots\frac{n-1}{(n-2)2}\frac n{(n-1)2} \\<\frac12\frac2{1\cdot2}\frac3{2\cdot2}\frac3{2\cdot2}\frac3{2\cdot2}\cdots\frac3{2\cdot2}\frac3{2\cdot2}\to0$$Possibly more obvious:$$n=1\frac21\frac32\frac43\frac54\cdots\frac{n-1}{n-2}\frac n{n-1}<1\frac21\frac32\frac32\...

1

If $a\le 0$ $$n\left( n^\frac 1n - 2^\frac 1n\right)^a = \frac{n}{\left(n^\frac 1n - 2^\frac 1n\right)^{|a|}} > \frac{n}{\left(n^\frac 1n \right)^{|a|}} = n^{1-\frac 1n |a|} \to \infty$$ If $a>0$ $$n^\frac 1n = e^{\frac 1n \ln n} = 1+ \frac 1n \ln n+O\left(\frac{\ln^2 n}{n^2}\right)$$ $$2^\frac 1n = e^{\frac 1n \ln 2} = 1+ \frac 1n \ln 2+O\left(\frac{1}... 0 Your proof seems legit. However, your problem can be easy solved using L'Hospital Rule. \lim\limits_{n\to\infty}\frac{n}{2^n}=\lim\limits_{n\to\infty}\frac{1}{2^n\log2}=0 1 If \;\lim\limits_{x \rightarrow10}g(x)=a\; and if \;\lim\limits_{x\rightarrow10}\dfrac{f(x)}{g(x)}=b\;,\; where \;a,b\neq 0\;,\; what's \;\lim\limits_{x\rightarrow10}f(x)\;? Since \;a\ne0\;,\; there exists \;\delta>0\; such that \;g(x)\ne0\; for any \;x\in\left]10-\delta,10+\delta\right[\;. Consequently, f(x)=g(x)\cdot\dfrac{f(x)}{g(x)}\... 0 Let f(0,0):=\lim f(x,y). If we look at the first limit along the path y=0, it evaluates to f(0,0). If we look at the first limit along the path y=x, it evaluates to \frac{f(0,0)-1}{1+f(0,0)}. If we look at the first limit along the path y=-x, it evaluates to \frac{f(0,0)+1}{1-f(0,0)}. If all these shall be equal, we must have f(0,0)=\pm i. 2 Usually, no, when we say a limit exists, that includes a requirement that it is finite. Such limits can, for instance, sensibly be used in further calculations, so it is a very natural distinction to make. Compared to a limit of \infty, which is still, in some sense, a definite answer to what the limit is, but it's not really a sensible result as an ... 0 Knowing that$$\begin{cases} \sum_{k=1}^n k &= \frac{n(n+1)}{2}\\ \sum_{k=1}^n k^2 &= \frac{n(n+1)(2n+1)}{6} \end{cases}$$You get$$\frac{A_n \cdot B_n}{\Vert A_n \Vert \Vert B_n \Vert} = \frac{n(n+1)/2}{\sqrt n \sqrt{n(n+1)(2n+1)/6}} \to \frac{\sqrt 3}{2}$$as n \to \infty and therefore$$\theta = \frac{\pi}{6}$$as \arccos is continuous. 1 You have$$\frac{x\sqrt{x}-3\sqrt{x-1}}{\sqrt{x}+2\sqrt{x(x^2+1)}} = \frac{x\sqrt{x}\left( 1- \frac{3}{x}\sqrt{1-\frac{1}{x}}\right)}{x\sqrt{x}\left( \frac{1}{x} + 2 \sqrt{1+\frac{1}{x^2}}\right)} =\frac{ 1- \frac{3}{x}\sqrt{1-\frac{1}{x}}}{\frac{1}{x} + 2 \sqrt{1+\frac{1}{x^2}}} $$and there is no more undeterminate form, so you have$$\lim_{x \rightarrow +\...

3

$f(x)= \frac{f(x)}{g(x)} \cdot g(x)$ for $x$ in a neighborhood of $10$, since $a \ne 0.$ Hence $$f(x)= \frac{f(x)}{g(x)} \cdot g(x) \to b \cdot a$$ as $x \to 10.$

0

Now I wasn't able to figure out a closed form but maybe it has something to do with the idea that I will present here. We will try different values for the parameter a. If a is equal to $0$ then obviously you just have the limit $\lim_{n\to\infty}n$, which diverges to $\infty$. If you let a be in the interval $(0;1]$ then it will diverge to $\infty$ once ...

0

This is merely an attempt to inspire, i was not able to conclude this way. So i think you can procede in this way. $$g_{c,d}(x)=\int_{\frac{1}{2}}^xz^{c-1}\sum_{i=0}^{d-1}{{d-1}\choose{i}}(-z)^{d-i-1}dz=$$ =\sum_{i=0}^{d-1}{{d-1}\choose{i}}\int_{\frac{1}{2}}^xz^{c-1}(-z)^{d-i-1}dz=\sum_{i=0}^{d-1}(-1)^{d-i-1}{{d-1}\choose{i}}\int_{\frac{1}{2}}^xz^{c+d-i-2}... 1 Since you are asking how to solve problems like this, I add this answer. Look carefully at the denominator: For y=-2x it is not defined while for x\neq 0 the numerator x^2-y^2 = x^2-(-2x)^2 = -3x^2\neq 0. This suggests that if you move to (0,0) along a curve close t0 y=-2x, you may push the value of the expression \frac{{x^2 - y^2}}{{2x + y}} to ... 0 \begin{align} \lim_{(x,y)\to(0,0)}\frac{{(x + y)\operatorname{tg}(x^2+y^2)}}{{\sqrt{x^2 + y^2}}} &= \left(\lim_{(x,y)\to(0,0)}(x + y)\right) \left(\lim_{(x,y)\to(0,0)}\frac{{\operatorname{tg}(x^2+y^2)}}{{\sqrt{x^2 + y^2}}}\right) \\ &= 0\cdot0 = 0 \end{align} by the hint from @JosÃ© Carlos Santos! 0 For (1): if you take \epsilon = 1/20<|x-1|<\delta_0 => |f(x)-2|< \frac{1}{2}$$does this \delta_0 exist? no becouse we could still take x=1+\frac{1}{n} for some n such that 1/n < \delta_0, so we have that |f(1+\frac{1}{n})-2|=|1+\frac{1}{n}-2|=|\frac{1}{n}-1|\ge \frac{1}{2} for n large enough. For (2): if we have that$$\forall \...

0

If $\epsilon=\dfrac12$, $$\exists\,\delta:\forall\,0<|x-1|<\delta\implies |x-2|<\dfrac12$$ is false.

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You have no control over $\frac y x$ so your answer is wrong. The expression is not defined along that lime $y=-2x$ so the limit is not even well defined. Also along the sequence $(\frac 1 n, \frac 1 {n^{2}}-\frac 2 n)$ you get a non-zero limit.

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For the case of sequences, the $\varepsilon-\delta$ type proof is as follows: Given $\varepsilon>0$, you need to find $N\in\Bbb N$ (depending on $\varepsilon$) such that for all $n\ge N$, it holds that $|s_n-L|<\varepsilon$ (in your case, $|s_n-1|$). For that purpose, observe that |s_n-1|=\left|\frac{n}{n+\sqrt n}-1\right|=\left|\frac {n-(n+\sqrt n)... 2 Observe that \begin{align*} \biggl|\frac{n}{n+\sqrt{n}} -1\biggr| = \biggl|\frac{\sqrt{n}}{n+\sqrt{n}}\biggr|\leq \frac{\sqrt{n}}{n}=\frac{1}{\sqrt{n}} \end{align*} Hence you have to choose an n such that \epsilon >\frac{1}{\sqrt{n}} happens. 0y=\dfrac{(1+x)^{\tfrac1x}-e}{x}z=(1+x)^{\tfrac1x}\implies \log(z)=\tfrac1x\log(1+x)$$Now Taylor$$\log(z)=\tfrac1x\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right) \right)=1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)z=e^{\log(z)}=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)y=-\frac{e}{2}+\frac{11 e x}{24}+O\left(x^2\right)$$0 Hospital: \lim_{x \rightarrow Ï€}\dfrac{n\cos (nx)}{m\cos(mx)}= (n/m)(-1)^n(-1)^m= (n/m)(-1)^{n+m}. 1 Let x-\pi=y$$\sin(mx)=\sin m(\pi+y)=\sin(my)\cos(m\pi)$$Now for integer m,\cos(m\pi)=(-1)^m For integer m,n;$$\lim_{x\to\pi}\dfrac{\sin mx}{\sin nx}=(-1)^{m-n}\dfrac mn\cdot\dfrac{\lim_{y\to0}\dfrac{\sin my}{my}}{\lim_{y\to0}\dfrac{\sin ny}{ny}}$$2 Without series, only L'Hospital$$ \lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x} $$we get$$ \lim_{x\to 0}\left[\left(-\frac{\ln(1+x)}{x^2}+\frac{1}{x}(1+x)^{-1}\right)(1+x)^{\frac{1}{x}}\right] \\ =\lim_{x\to 0}\left(-\frac{\ln(1+x)}{x^2}+\frac{\frac{1}{1+x}}{x}\right)\lim_{x\to 0}(1+x)^{\frac{1}{x}} \\ =\lim_{x\to 0}\left(\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}\...

0

Yeah probably write (1+x)1/x can be written as e(1/x)ln(1+x) so the final expression looks like (e/x)e(1/x)ln(1+x) - 1 so now you can solve it using ln(1+x) expansion i.e. x - x2/2 + x3/3 +.... so you will be getting the answer without L'Hospital rule

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Hint Taylor series of $${(1+x)}^{1/x}=e(1-\frac{1}{2}x+\frac{11}{24}x^2+...)$$ If used you get $-e/2$

3

\begin{aligned} &\lim_{x\to 0}\frac{(1+x)^{\frac{1}{x}}-e}{x}\\ &=\lim_{x\to 0}\frac{e^{{\frac{1}{x}}\ln(1+x)}-e}{x}\\ &=e\lim_{x\to 0}\frac{e^{{\frac{1}{x}}\ln(1+x)-1}-1}{x}\\ &=e\lim_{x\to 0}\frac{{\frac{1}{x}}\ln(1+x)-1}{x}\\ &=e\lim_{x\to 0}\frac{\ln(1+x)-x}{x^2}\\ &=e\lim_{x\to 0}\frac{\frac{1}{1+x}-1}{2x}\\ &=e\lim_{x\to 0}\...

1

I think this follows directly from the definition of the derivative: $$f'(z) = \lim_{z \rightarrow z_0} \frac{f(z)-f(z_0)}{z-z_0} = \lim_{z \rightarrow z_0} \frac{f(z)}{z-z_0}$$

0

\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\... 3 The expression in the square root needs to be non-negative: $$5x-4-x^2=\tfrac94-(x-\tfrac52)^2\geq0$$ $$(x-\tfrac52)^2\leq\tfrac94$$ $$|x-\tfrac52|\leq\tfrac32$$ $$-\tfrac32\leq x-\tfrac52\leq\tfrac32$$ $$1\leq x\leq4$$ On the other hand, the denominator needs to be non-zero: $$\lnot\big[\,\lfloor x\rfloor-3=0\,\big]$$ $$\lnot\big[\,3\leq x<4\,\big]$$ $$x&... 0 f is continuous on all intervals \left(\frac{1}{n + 1}, \frac{1}{n}\right), n \gt 0 and \left(\frac{1}{n}, \frac{1}{n - 1}\right), n \lt 0. f is discontinuous at all points \frac{1}{n}, n \in \mathbb{Z} because f(\frac{1}{n}) = 1 \neq 0 = \lim_{x \to \frac{1}{n}} f(x). These are all removable discontinuities. The only other point of interest is ... 2 Anyway, no need for L'Hospital's rule here. Just use Maclaurin's expansions at the relevant orders and compose them where required: \mathrm e^x=1+x+o(x); \sin^2x=x^2+o(x^2), so \mathrm e^{\sin^2x}=1+x^2+o(x^2)=1+o(x); \sin 2x=2x+o(x). Therefore$$\frac{\mathrm e^{\sin^2x}-\mathrm e^x}{ {\sin2x}}=\frac{-x+o(x)}{2x+o(x)}=-\frac12+o(1).$$5 Just to give an alternative to J.G.'s approach, we have$$\begin{align} {e^{\sin^2x}-e^x\over\sin2x} &={(e^{\sin^2x}-1)-(e^x-1)\over2\sin x\cos x}\\ &={1\over2\cos x}\left({e^{\sin^2x}-1\over\sin^2x}\sin x-{e^x-1\over x}\cdot{x\over\sin x} \right)\\ &\to{1\over2\cdot1}\left(1\cdot0-1\cdot1 \right)\\ &=-{1\over2} \end{align}$$9 You want$$\underbrace{\lim_{x\to0}e^x}_{1}\cdot\underbrace{\lim_{x\to0}\frac{e^{\sin^2x-x}-1}{\sin^2x-x}}_{1}\cdot\lim_{x\to0}\frac{\sin^2x-x}{\sin(2x)}$$(the second limit uses \lim_{y\to0}\frac{e^y-1}{y}=1). The last limit is$$\underbrace{\lim_{x\to0}\frac12\tan x}_0-\underbrace{\lim_{x\to0}\frac{x}{\sin(2x)}}_{1/2}=-\frac12.$1 You can simply use the equivalence of the functions sinus and exponential at$x=0$. Thus$\sin(x)$is equivalent to$x$at$x=0$, and$(\sin(x))^2$is equivalent to$x^2$at$x=0$. In addition,$\exp(x)$is equivalent to$1+x$at$x=0$. Bearing this in mind, you find that$\exp( (\sin(x))^2 ) - \exp(x)$is equivalent to$(1+x^2) - (1+x) = x(x-1)$at$x=0\$. ...

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