New answers tagged

1

You can prove that the limsup is 1 by showing that for any $k\in\mathbb{N}$ $\exists n_k>k$ with $ |\sin (n_k\theta)|>0.5$. Now take the increasing series $(n_k)_{k=1}^{\infty},$ with $n_k$ beeing the smallest integer $\geq k$ such that $ |\sin (n_k\theta)|>0.5.$ It now follows that $$ 1\geq \lim_{k\to\infty}|\sin (n_k\theta)|^{1/n_k}\geq \lim_{k\...


0

Take $$c_n=n^2+2$$ then $$\forall n \;\; c_n>1$$ and $$u_n=\frac{1}{c_n}-\frac{1}{c_nc_{n+1}}=$$ $$\frac{1}{c_n}(1-\frac{1}{c_{n+1}})\sim \frac{1}{c_n}$$ since $\lim_{n\to+\infty}c_n=+\infty$. we conclude that the infinite sum $$\sum_{n=0}^{\infty}u_n \text{ is finite}$$ cause the series $\sum \frac{1}{n^2+2}$ is convergent.


0

For any $z$ you can define $E(z)=\lim_{n\to\infty}\left(1+\frac zn\right)^n$. The cited argument shows that $E(z)=\exp(z)$ for positive $z$, where $\exp$ is the inverse function of $\ln$. In the cited pages it is not shown that $\exp$ is an exponential function, so that writing $\exp(z)=e^z$ is not really justified. For that one would need to show that the ...


0

Besides taking the limit of a function, you can take the limit of any relation, thought of as a multi-valued function. Recall that $ \lim _ { x \to x _ 0 } y = L $, where $ y = f ( x ) $ for some function $ f $, means that there is a unique $ L $ such that, for each $ \epsilon > 0 $, for some $ \delta > 0 $, for each $ x $ in the domain of $ f $, if $ ...


0

Notice that $$1-\frac1n=\frac1{\dfrac n{n-1}}=\frac1{1+\dfrac1{n-1}}.$$ Then raising to the power $n$ or $n-1$ makes no difference (the factor tends to $1$) and $$f(-1,n)=\frac1{f(1,n)}$$ in the limit.


2

Note that $\log\lvert x\rvert$ is differentiable in $\mathbb R\setminus\{0\}$ and if you differentiate it, you get $\frac1x$. So, yes, you can take negative values for $x$.


3

This is false as stated (probably it's true with some reasonable additional hypothesis, maybe regarding monotonicity of $f'$). The Idea: We take $f(x)=\int_0^x f'(t)\,dt$, where $f'>0$ is constructed so that $f'(n)\to\infty$, but $f'$ is very small except at points very close to a positive integer, so that $\int_0^\infty f'\le1$. Then $f(x)\le1$, so the ...


0

The numerator is $\int_0^1 x\log(x)dx=-1/4$ so $\lim_{n\to\infty}\frac{1}{n}(\sum_{k=1}^n \frac{2k-1}{2n-1}\log(\frac{2k-1}{2n-1}))=-\frac14$ so $\lim_{n\to\infty}(\prod_{k=1}^n(\frac{2k-1}{2n-1})^{2k-1})^{\frac{1}{n(2n-1)}}=e^{-\frac14}$


1

Using Stirling formula $\sqrt{2\pi}n^{n+\frac12}e^{-n}\leq n! \leq \sqrt{2\pi}n^{n+\frac12}e^{-n+\frac1{12n}}$ and $(2n-1)!!=\frac{(2n-1)!}{(2n-2)!}$


0

$$ \frac {1}{x^2}-\frac{1}{\sin^2x} = \left(\frac {1}{x}+\frac{1}{\sin x}\right)\left(\frac {1}{x}-\frac{1}{\sin x}\right) = \frac{1}{x^2}\left(\frac {x}{x}+\frac{x}{\sin x}\right)\left(\frac {x}{x}-\frac{x}{\sin x}\right) $$ so $$ \lim_{x\to 0}\left(\frac {1}{x^2}-\frac{1}{\sin^2x}\right) =2\lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{x}{\sin x}\right) = 2\...


0

I'll try to recreate the steps. Step 1): We are subtracting two fractions so let's have a common denominator, leading to $$\lim_{x \rightarrow 0}{\sin^2x - x^2 \over x^2\sin^2 x}$$ Step 2): I see a difference of two squares so let's factor: $$\lim_{x \rightarrow 0}{(\sin x - x)(\sin x + x) \over x^2 \sin^2 x}$$ Step 3): I know the limit of ${\sin x \over ...


0

Since nobody has explicitly mentioned it, I am going to state for the record that the best way to evaluate limits in general is not actually via purely algebraic manipulation of the kind you are looking for, but rather via asymptotic expansions (which is how all computer algebra systems do it): $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $ As $x → 0$:   $\...


2

As $x \to -\infty$ and $y \to -\infty$ is is clear that the limit is $0$, not $\infty$. If you are considering only positive values of $x$ and $y$ then the inequalities $e^{x} > \frac {x^{2}} 2$ and $e^{y} > \frac {y^{2}} 2$ make the result quite evident.


1

For $x$ and $y$ large enough you have that $e^{|x|}>x^2$ and $e^{|y|}>y^2$ so $e^{|x|}+e^{|y|}>x^2+y^2=||(x,y)||^2$ that means $\frac{e^{|x|}+e^{|y|}}{||(x,y)||}>||(x,y)||\to \infty$ In your case you have a problem because for $x\to-\infty$ and $y\to-\infty$ you have that the limit is zero.


0

The numerator is like $x$, since all the functions are $x + O(x^3)$, and the denominator is like $5x$ since $\sqrt{x}\sin(3x)$ is like $3x^{3/2}$ and this and $x^2$ are smaller than $x$. The result is therefore $\dfrac15$.


0

Hint $$\sin(\arctan(\sin x))=\frac{\sin (x)}{\sqrt{1+\sin ^2(x)}}$$ makes the problem much simpler. Concerning the denominator, it behaves like $k x$ (I let you finding $k$).


5

OK, let's talk through the thought process. When I see a difference of two fractions, I give them common denominators first, as per your second $=$. I can't help but factorise the new numerator's difference of two squares after that. Since there's a division by $\sin^2 x$, which $\to0$ as $x\to0$, I need to take out a $\left(\frac{\sin x}{x}\right)^{-2}$ ...


2

Just "gets closer" is not enough, because it might not get arbitrarily close: consider for example $f(x)=x^2,c=0,L=-1$. On the flip side, even if $\lim_{x \to c} f(x)=L$, you still might not have $f(x)$ getting closer to $L$ as $x \to c$ if in fact $f(x)=L$ for various $x$'s arbitrarily close to $c$. For example you could have $f(x)$ is equal to $L$ ...


3

One could also write directly: $$0\leq \frac{|x|}{\sqrt{x^{4}+4x^{2}+7}}\leq \frac{|x|}{\sqrt{7}},$$ as $$ \sqrt{x^{4}+4x^{2}+7} \geq 7.$$


4

Once you know that $\dfrac{\sin x}{x} \to 1$, and similar, fairly simple limits, it becomes natural to try to make them appear by extracting them from more complicated expressions.


7

Since$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots,$$you know that$$\sin(x)+x=2x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$and that$$\sin(x)-x=-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots.$$Therefore both limits$$\lim_{x\to0}\frac{\sin(x)+x}x\text{ and }\lim_{x\to0}\frac{\sin(x)-x}{x^3}$$exist; they are equal to $2$ and to $-\frac16$ respectively. This explains why ...


3

Hint for $|x|\to +\infty$: note that for $x\not=0$, $$0\leq \frac{|x|}{\sqrt{x^{4}+4x^{2}+7}}\leq \frac{|x|}{\sqrt{x^{4}}}.$$ Hint for $x\to 0$: we have that $$0\leq \frac{|x|}{\sqrt{x^{4}+4x^{2}+7}}\leq |x|.$$


1

$$\frac{(2m)!}{m!^2}={2m \choose m} \ge \prod_{m < p \le 2m} p, \qquad {2^{k+1} \choose 2^k}\le 8 {2^k \choose 2^{k-1}}, \qquad \prod_{p \le n} p \le \prod_{k \le \log_2(n)+1} {2^k \choose 2^{k-1}} \le 4^{4n}$$ If $\pi(n) \ge Cn$ then $\prod_{p \le n} p \ge q^{Cn-q}$ thus for $n$ large enough we must have $\pi(n) < Cn$ which means $$\lim_{n \to \infty}...


2

If the limit exists and is $L$, then $\sum_{k=1}^n kL \approx \prod_{k=1}^n L^k $ or $Ln(n+1)/2 \approx L^{n(n+1)/2} $ or $L \approx (n(n+1)/2)^{1/((n(n+1)/2-1)} $ and since $n^{1/n} \to 1$, $L = 1$.


3

Let $p(x) = \frac{x}{\ln x}$, being the approximate prime counting function. That means there are approximately $\frac{x}{\ln x}$ primes less than or equal to x, and $x-\frac{x}{\ln x}$ composites less or equal to than x. First, let's derive $p_n$. Since there are about $\frac{x}{\ln x}$ primes less than or equal to x, there are x primes less than or equal ...


1

If some series approaches $\infty$, it is pretty standard to say that series has no limit, for this exact reason. The limit is unbounded. $\infty$ does not exist, so it makes little sense to say it actually equals infinity. I've written before that $$\lim_{x\to0}\frac{1}{x}=\infty$$ but this is more of a shorthand or even an aesthetical choice than it is ...


0

Let $y = \displaystyle \left( \frac{2}{e} \right)^{2n}\sqrt{n}$ or, $\displaystyle y = \left(1 - k \right)^{2n} \sqrt{n}$ where $k = \frac{e-2}{e} < 1$ or, $\displaystyle y = \frac{\sqrt{n}}{\left(1 - k \right)^{-2n} }$ Now apply L'Hospital's rule: $\displaystyle \lim_{n \to \infty} y $ $\displaystyle = \lim_{n \to \infty} \frac{\sqrt{n}}{\left(1 - ...


0

Hint: If $a>1$, $\; n^r=_{\infty}o(a^n)$ for any $r\in\mathbf R$.


0

Write your limit as $$e^{\lim_{n\to\infty}(2n(\ln(2)-1)+\frac{1}{2}\ln(n)}$$


1

Try $y_n=n$. We have $$\begin{align} \lim_{n\to\infty}{u_n\over y_n}&=\lim_{n\to\infty}{(n+1)^{1/n+1} - n^{1/n}\over n}\\&=\lim_{n\to\infty}{n+1\over n}(n+1)^{\frac1n}-{n^{1/n}\over n}\\ &= 1\cdot 1-0\\&=1 \end{align}$$


1

$$(n+1)^{1/(n+1)}-n^{1/n}=\sqrt[n]{n}\left(\frac{(n+1)^{1/(n+1)}}{n^{1/n}}-1\right)$$ Take $v_n=\left(\frac{(n+1)^{1/(n+1)}}{n^{1/n}}-1\right)$. Then $\frac{u_n}{v_n}=\sqrt[n]{n}\to1$.


1

We have \begin{align} u_n &= \exp(\frac{\ln(1+n)}{1+n}) - \exp(\frac{\ln(n)}{n})\\ &= \exp(\frac{\ln(1+n)}{1+n})(1 - \exp(\frac{\ln(n)}{n}-\frac{\ln(1+n)}{1+n}))\\ \end{align} With $$ 1 - \exp(\frac{\ln(n)}{n}-\frac{\ln(1+n)}{1+n}) \sim \frac{\ln(1+n)}{1+n} - \frac{\ln(n)}{n} $$ and $$ \exp(\frac{\ln(1+n)}{1+n}) \sim 1 $$ we get $$ u_n \sim \frac{\...


9

$\lim_{n\rightarrow \infty}x_n = 1$. Since $x_n\geq1,\forall n\geq 1$, we have $x^n_n\geq x_n$, hence $\lim x_n^n\geq \lim x_n \geq 1$, which implies $\lim x_n=1$.


5

There is too much of unnecessary hypothesis in this question. Let $\{x_n\}$ be any sequence of real numbers such that $x_n^{n} \to 1$. Then $x_n >0$ after some stage . Taking logarithm we get $nlog \, x_n \to 0$. Since $\log x_n =\frac 1 n log x_n$ we get $\lim \log \, x_n=0$ or $\lim x_n=1$.


3

As $\lim_{n \to \infty} x_n^n = 1$, we have $x_n^n < 2$ for large enough $n$, so $x_n < \sqrt[n]{2}$. Also $x_n \geqslant 1$. As $\sqrt[n]{2} \to 1$ and $1 \to 1$, by squeeze theorem we have $x_n \to 1$.


1

I've had this same question myself. As I understand it, the limit is the real number $L$ that $f(x)$ get's $\mathit{most}$ $\mathit{arbitrarily}$ $\mathit{close}$ to as $x$ approaches $a$, i.e. there does not exist any other real number that $f(x)$ gets closer to. The limit is not making a statement about the "true" value of $f(x)$ that is achieved when $...


0

I would write \begin{align} \int_{r_1}^{r_2} p(x)dx + \int_{r_1}^{r_2} \frac{1}{x} dx = \Big[P(x)\Big]_{r_1}^{r_2} + \Big[\ln|x|\Big]_{r_1}^{r_2} = \Big(P(r_2) - P(r_1)\Big) + \Big(\ln|r_2| - \ln|r_1| \Big) \end{align} and make the following observations In general, people typically write $\int f(x)dx=F(x)+C$ where $(F(x)+C)$ is the general ...


0

This is not true. If $\alpha=\beta =\gamma=1$ the the function is identically $1$ so the limit is $1$.


2

There's no need for L'Hopital, just a change of variable to make the limit easier to see. If you let $x=e^u$ with $u\to\infty$, then, with $\alpha,\beta,\gamma\gt0$, we have $$x^\alpha e^{-\beta\ln^\gamma x}=e^{\alpha u-bu^\gamma}\to \begin{cases} 0\quad\text{if }\gamma\gt1\text{ or }\gamma=1\text{ and }\alpha\lt\beta\\ 1\quad\text{if }\gamma=1\text{ and }\...


1

The useful trick is to use the fact that if $\lim_{x \to \infty} f(x) \geq 0$ then it is equal to $\exp \left ( \lim_{x \to \infty} \log(f(x)) \right )$, with the understanding $e^{-\infty}=0$. So you have $\exp \left ( \lim_{x \to \infty} \alpha \log(x) - \beta \log(x)^\gamma \right )$. Now you can change variables in the limiting procedure to $u=\log(x)$, ...


0

You can simplify it: $$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}=\lim_{x \to 0}\frac{x+2-\sqrt{2}\sqrt{x+2}}{3(x+1)+\sqrt{x+1}-4}=\\ \lim_{x \to 0}\frac{\sqrt{x+2}(\sqrt{x+2}-\sqrt{2})}{3(\sqrt{x+1}-1)(\sqrt{x+1}+\frac43)}=\\ \lim_{x \to 0}\frac{\sqrt{x+2}-\sqrt{2}}{\sqrt{x+1}-1}\cdot \lim_\limits{x\to 0}\frac{\sqrt{x+2}}{3\sqrt{x+1}+4}=\\ \frac{\...


-1

For any $\epsilon > 0$ let $\delta = 5.6784\times 10^{237}$. If $|x - \pi| < \delta$ then $|3-3| = 0 < \epsilon$. So $\lim_{x\to \pi} 3 = 3$ ===== This follows the definition precisely. The definition is literally that $\lim_{x\to a} f(x) = L$ if for any $\epsilon >0$ we can find a $\delta$ (based upon $\epsilon$) so that whenever $|x-a|&...


2

The "$\varepsilon$ - $\delta$" definition of a limit is the following: We say that $L$ is the limit of a function $f$ at $c$ (i.e., we say that $\lim_{x \to c} f(x) = L$) if, for every $\varepsilon > 0$ there exists some $\delta > 0$ such that, whenever $|x-c| < \delta$, we have $|f(x) - L| < \varepsilon$. As stated in the comments, having a ...


1

$$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}=\lim_{x\rightarrow0}\frac{((x+2)^2-(2x+4))(3x-1-\sqrt{x+1})}{((3x-1)^2-(x+1))(x+2+\sqrt{2x+4})}=$$ $$=\lim_{x\rightarrow0}\frac{(x^2+2x)(3x-1-\sqrt{x+1})}{(9x^2-7x)(x+2+\sqrt{2x+4})}=\lim_{x\rightarrow0}\frac{(x+2)(3x-1-\sqrt{x+1})}{(9x-7)(x+2+\sqrt{2x+4})}=\frac{2\cdot(-2)}{-7\cdot4}=\frac{1}{7}.$$


0

As evaluating the limit as $x\to 0$ forms $$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}=\frac{2-\sqrt{4}}{-1+\sqrt{1}}=\frac{0}{0}$$ and $\frac{0}{0}$ is an indeterminate form, apply L'Hopital's rule by taking the derivative of the numerator and denominator $$\lim_{x \to 0}\frac{1-\frac{1}{\sqrt{2x+4}}}{3+\frac{1}{2\sqrt{x+1}}}=\frac{1-\frac{1}{...


2

You may use $$\sqrt{1+x}=1+\frac x2+O(x^2)$$ First, $$\sqrt{2x+4}=2\sqrt{1+\frac x2}=2(1+\frac x4)+O(x^2)=2+\frac x2+O(x^2)$$ And $$\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}} =\frac{x+2-2-\frac x2+O(x^2)}{3x-1+1+\frac x2+O(x^2)} \\=\frac{\frac x2+O(x^2)}{\frac72x+O(x^2)}=\frac17+O(x)\underset{x\to0}\longrightarrow\frac17 $$


1

Basically you have$\lim\limits_{n \to \infty} \frac{1}n f(n)=c$. If you calculate the limit w.r.t $n$ it is not allowed that $n$ appears on the RHS. $$\lim\limits_{n \to \infty} f(n)\neq n\cdot c$$


1

No. Let $t_n = \frac{1}{n}\sum_{i=1}^{n}(\frac{n}{n-i})^{\gamma}$. Then, by your hypothesis, $\lim_{n \to \infty} t_n = c$ for some $c \in \mathbb R$. Most likely, for some large $N$, $$ t_N < t_{N+1} < t_{N+2} < \cdots < c$$


4

Notice that taking the limit as $r \to 0$, like you did, only gets near zero through paths that are straight lines. This is not enough to ensure differentiability and this example is to show why. As pointed out in the comments, we have that $$ \frac{\cos^3 \theta + \sin^3 \theta}{\cos \theta - \sin \theta} $$ is not bounded, when you approach zero through ...


3

Your mistake is in falsely assuming that $g(\theta) = \dfrac{\cos^3\theta+\sin^3\theta}{\cos\theta-\sin\theta}$ is bounded. It has a discontinuity at $\theta = \pi/4$ so it is not bounded. Hence you cannot conclude $ \lim\limits_{r\to 0}f(r) g(\theta)=0$


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