New answers tagged limits
1
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Trying to calculate a limit with a finite product and WolframAlpha disagrees with my logic.
Typing in Wolfram Alpha the command
FunctionExpand[Product[2j/(N+j+1),{j,1,N}]]
it properly returns $$\frac{\sqrt{\pi }\,\, 2^{-N-1}\,\, \Gamma (N+2)}{\Gamma
\...
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5
votes
Accepted
Trying to calculate a limit with a finite product and WolframAlpha disagrees with my logic.
You're right and WolframAlpha is wrong.
Note that the first factor is $\frac2{N+2}$ and all the other factors (as you noted) are less than $1$. So the product is actually less than $\frac2{N+2}$, ...
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Proof a sequence is convergent
Observe ($\sqrt{\cdot}$ is cubic root),
$$ y^2_n \leq \frac{1}{n}+ \frac{|x_n \sqrt{x_n}|}{2}(y_n^2+1) \implies y_n^2\left(1 -\frac{|x_n \sqrt{x_n}|}{2}\right) \leq \frac{1}{n} + \frac{|x_n \sqrt{x_n}...
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Limit of $f(x,y)$ when $x^2+y^2$ tends to $\infty$
obvious ho to get large positive values, say $x=y $ or $x=7y.$
Next, given an $x,y$ pair you like, what is
$u^2 + 3uv + v^2$ when
$$ u = 5x + 2y, \; \; \; v = -2 x - y \; \; \; ? \; \; $$
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2
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Limit of $f(x,y)$ when $x^2+y^2$ tends to $\infty$
There's no limit as $x^2+y^2\rightarrow \infty$.
Consider the sequence $(x_n,y_n) = (n,n)$, then $x_n^2+y_n^2= 2n^2$ converges to infinity as $n\rightarrow \infty$ while $f(x_n,y_n) = 5n^2$ and $\lim_{...
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2
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how to prove that $\displaystyle \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}= \infty$ without using Stirling's approximation
We have $$4^n=2^{2n}>{2n\choose n}={(2n)!\over ( n!)^2}$$ Hence $$a_n={4^n(n!)^2\over (2n)!}\ge 1$$ so the series $\sum a_n$ cannot be convergent. Moreover the sequence $a_n$ tends to infinity at ...
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0
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Can infinite summation be defined without using limits?
In your comments, you said you consider the least-upper bound property to be "simpler than the definition of a limit".
These two concepts are closer to each other than perhaps you might ...
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2
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Accepted
how to prove that $\displaystyle \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}= \infty$ without using Stirling's approximation
Use the Quotient-Criterion in a bit more specific form.
With $a_n = \frac{(n!)^2 4^n}{(2n)!}$ you get
$$\frac{a_{n+1}}{a_n}= 2\frac{n+1}{2n+1}=1+\frac 1{2n+1}$$
Hence, with $a_1=2$ we get using a ...
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3
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how to prove that $\displaystyle \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}= \infty$ without using Stirling's approximation
First we split $(2n)!$ into product of even factors $\cdot$ product of odd factors i.e. : $(2n)! = (2n)(2n-2)..(2) \cdot (2n-1)(2n-3)..(1)$
Now we write $4^n = (2^2)^n = 2^{2n} = 2^n \cdot 2^n$
We ...
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2
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how to prove that $\displaystyle \lim_{n \to \infty } \frac{(n!)^24^n}{(2n)!}= \infty$ without using Stirling's approximation
Hint: Use $(2n)!=(2n)!!(2n-1)!!<(2n)!!^2$.
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1
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Let $f(x) = \tanh(x) - \frac{1}{x}$ find $\lim_{t \to 0} t f^{-1}(t) $
Let $x=f^{-1}(\sqrt{1-t}).$ Then
$t=1-f(x)^2.$ When $t\to 0, $ then $\sqrt{1-t}\to 1,$ and consequently $x\to \infty .$
Therefore
$$\lim_{t\to 0}tf^{-1}(\sqrt{1-t})=\lim_{x\to \infty }[1-f(x)^2]\,x=
\...
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-1
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Show that $\lim_{x\rightarrow 2}\frac{x^2-4}{x-2}=4$.
$$\lim_{x\to 2} \frac{x^2-4}{x-2}=\lim_{x\to 2} \frac{(x-2)(x+2)}{x-2}=\lim_{x\to 2} \frac{x+2}{1}=2+2=4\square$$ Or by L'hopital:
$$\lim_{x\to 2} \frac{x^2-4}{x-2}=\lim_{x\to 2} \frac{2x}{1}=2\cdot 2=...
2
votes
Accepted
Show that $\lim_{x\rightarrow 2}\frac{x^2-4}{x-2}=4$.
You need to prove
$\forall \varepsilon > 0 \hspace{5mm} \exists \delta >0 \hspace{5mm} \forall x(d_x(x,p)<\delta \Rightarrow d_Y(f(x),q)< \varepsilon) \\ \forall \varepsilon > 0 \...
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1
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Doubts: continuity VS existence of the limit
Generally speaking, a textbook about continuous real functions will proceed in this order:
First, define continuity at a point contained in the domain of the function.
Then, define left-continuity at ...
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2
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Proof a sequence is convergent
For every $n \geq 0$, let $P_n : X \mapsto X^2 - Xx_n \sqrt[3]{x_n} - \dfrac{1}{n}$.
The hypothesis is that $P_n(y_n) \leq 0$ for every $n$.
But the roots of $P_n$ are exactly $\dfrac{x_n\sqrt[3]{x_n} ...
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2
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Find the exact value: $ \lim _ {n\rightarrow \infty } $ $ \sum _ {k=1}^ {n} $ $ \frac {n}{3n^ {2}+2kn+k^ {2}} $
$$\lim_{n\to\infty}\sum_{k=1}^n\frac {n}{k^2+2kn+3n^{2}}=
\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n
\frac{1}{3+2\frac{k}{n}+\frac{k^2}{n^2}}\\
=\int_{0}^{1}\frac{dx}{3+2x+x^2}
=\frac{1}{\sqrt{2}}\cot ^{...
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4
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Find the exact value: $ \lim _ {n\rightarrow \infty } $ $ \sum _ {k=1}^ {n} $ $ \frac {n}{3n^ {2}+2kn+k^ {2}} $
$$\frac {n}{k^2+2kn+3n^ {2}}=\frac {n}{(k-a)(k-b)}$$ where
$$(a,b)=-n\pm i n\sqrt{2}$$ Using partial fraction decomposition
$$\frac {n}{(k-a)(k-b)}=\frac n {a-b}\left(\frac{1}{k-a}-\frac{1}{k-b}\...
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5
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Find $\lim_{n\to\infty}\left(\int _{0}^{n}\frac{\mathrm dx}{x^{x^n}}\right)^n$
The analysis below is morally similar to Dominik's answer (+1 of course), but simplified (or at least simplified compared to the version of the answer I had seen when writing this, I understand it has ...
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0
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Find the exact value: $ \lim _ {n\rightarrow \infty } $ $ \sum _ {k=1}^ {n} $ $ \frac {n}{3n^ {2}+2kn+k^ {2}} $
This can be done with Mathematica without any integral sums in such a way.
Sum[n/(3 n^2 + 2*k*n + k^2), {k, 1, n}]
...
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0
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How to prove that a form like $0^\infty$ is not an indeterminate form? What makes a form indeterminate?
a op b is called an "indeterminate form" if you can't find the limit of f(x) op g(x) by knowing that the limit of f(x) is a and the limit of g(x) is b. And that is the case here: If you know ...
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How do I understand this solution for finding $\lim_{n\to\infty} {1/n}^{1/n}$?
The negative sign comes from $\ln(1/n)=-\ln(n).$
The final derivation comes from the continuity of the exponential function: if $\lim_{n\to\infty}a_n=0$ then $\lim_{n\to\infty}e^{a_n}=e^0$ (here, $a_n=...
3
votes
Accepted
Absolute convergence on boundary implies continuity of power series
As requested, let's make my comment as an answer. Note that if we consider $f_N(z)=\sum_{n=0}^{N}c_nz^n$ we have by the triangle inequality $|f_N(z)-f(z)| \le \sum_{n \ge N+1}|c_n|$ whenever $|z| \le ...
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2
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Doubts: continuity VS existence of the limit
For points in the interior of the domain, you need to distinguish left from right continuity (i.e. depending on the side the limit is taken from). There, a function is continuous if it is both left ...
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9
votes
Accepted
Doubts: continuity VS existence of the limit
Not an expert in real analysis here, but when checking for continuity you only approach the function from where it exists.
Let $f$ be a function whose domain and range are sets of real numbers. Let $...
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3
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Doubts: continuity VS existence of the limit
The domain of the square root is $[0,∞)$, so you cannot approach $0$ from below. It therefore suffices to approach $0$ from above! The limit should be the same no matter how one approaches $0$, and in ...
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1
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Solutions to $x^x=1$?
Your question is $x^{x}=1$. Now clearly $x=1$ satisfies the above equation because $1^{1}=1$. Again you can solve this above equation by taking $log$ on both sides.
$x^{x}=1$. Therefore $xln(x)=ln(1)$....
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Proof of Convergence: Babylonian Method $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$
An easy way to prove the Babylonian method works is to think of it as like binary search, but faster.
With each $x_n$, associate the closed interval $I_n$ between $x_n$ and $\frac {a}{x_n}$, inclusive....
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4
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Accepted
On kernels and stalks of sheaves.
For 1., there is indeed a general approach: given a category $\mathcal{C}$ and a (full) reflective subcategory $\mathcal{D}$ of it (i.e. the inclusion $i\colon \mathcal{D}\to\mathcal{C}$ is fully ...
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Find $\lim_{n\to\infty}\left(\int _{0}^{n}\frac{\mathrm dx}{x^{x^n}}\right)^n$
Let $f_n(x)=x^n\log x$. It is not difficult to show that
$$I_n=\int^n_0e^{-f_n}\xrightarrow{n\rightarrow\infty}1$$
This may be achieved by analyzing the integral over $(0,1]$ and $(1,\infty)$ ...
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1
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nth-derivative of a function of n
Let us calculate the Taylor series around $z=i$ of
$$g(z):=\frac{e^{iz}}{(z+i)^n}.$$
For $|h|<2,$
$$\begin{align}g(i+h)&=\frac{e^{i(i+h)}}{(2i+h)^n}\\
&=\frac{e^{-1}}{(2i)^n}\,\frac{e^{ih}}{...
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1
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Hello everyone, I want to ask one calculation, I tried my best , but still cannot find up answer , please help
You can use a small angle approximation
$$\lim_{x \rightarrow 0} \frac{\arcsin 5x}{\sin 3x} \approx \frac{5 x}{3 x} = \frac{5}{3} $$
- 13.2k
6
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Find $\lim_{n\to\infty}\left(\int _{0}^{n}\frac{\mathrm dx}{x^{x^n}}\right)^n$
Clearly, $\lim_{n \to \infty} \frac{1}{x^{x^n}} = 1_{(0,1]}(x)$ for any $x>0$. Moreover, $\frac{1}{x^{x^n}} \le 1_{[0,1]}(x) + \frac{1}{x^x} 1_{(1,\infty)}(x)$ and the latter is integrable, hence $...
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Trigonometric substitution in $\int \sqrt{\frac{x+7}{x}}\ dx$
Yes that would work fine.
Alternatively you can also try substituting $ x = t^2$ to get,
$$\int \sqrt{\frac{7+x}{x}}\ dx = \int\sqrt{\frac{7+t^2}{t^2}}\ 2t\ dt = 2\int\sqrt{7 + t^2}\ dt$$
Which is not ...
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if $\lim\limits_{n \to \infty} b_n =0 $ then how to prove that $\lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k}=0$
I made a mistake that is $b_n \to 0$ is true but there is another condition $$s_n :=\sum _{k=1}^n b_k = \sum _{k=1}^n a_{k+1}- a_k= a_{n+1 }- a_1$$
it is clear that $s_n \to a -a_1 $ i.e $\...
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1
vote
Accepted
lim sup of maximum
First, note that $f(n)g(n)\to 0 \implies \limsup_{n \to \infty} ( f(n) g(n) ) = 0.$
Here is a counter-example to the proposition in the question:
$
f(n)=
\begin{cases}
\log(n)&\text{if}\, n\text{ ...
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0
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Asymptotic of $\sum_{k=0}^\infty\frac{\Gamma (k+n+1) \Gamma (3 k+n+1)}{\Gamma (k+2) \Gamma (3 k+2 n+2)}$ as $n\to \infty$
If you simplify the hypergeometric function
$$f(n)=\frac{\sqrt \pi}{4^{n-1} }\,\frac{\Gamma (n-2)}{\Gamma \left(n-\frac{1}{2}\right)} \,(A-1)$$ where
$$A=\,
_4F_3\left(\frac{n-2}{3},\frac{n-1}{3},\...
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2
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Compute $\lim_{n\to\infty}(\frac{a_{n+1}}{\sqrt[n+1]{b_{n+1}}}-\frac{a_n}{\sqrt[n]{b_n}})$ given $(a_{n+1}-a_n)/n\to a$ and $b_{n+1}/(nb_n)\to b$
Here is a rather nice result which helps in this kinds of problems:
Theorem (O. Carja): Suppose $(a_n:n\in\mathbb{N})$ is a sequence of positive numbers such that
(i) $\lim_n\frac{p_n}{n}=p>0$
(...
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2
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Define $f(0,0)$ so that $f'_x(0,0)$ exists
You seem to have assumed that $ f ( 0 , 0 ) $ is $ 0 $, and that didn't work. Since you don't know what $ f ( 0 , 0 ) $ should be, just write it as $ c $ or something. So you have $$ \lim _ { h \to ...
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6
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Accepted
What's the limit of the sequence $a_{n} = \frac{\lfloor\sqrt{2n}\rfloor}{\lfloor\sqrt{n}\rfloor}$?
Hint: using the fact that, for all $x$, we have $\lfloor x\in (x-1, x]$, we have $$a_n = \frac{\lfloor\sqrt{2n}\rfloor}{\lfloor n\rfloor} \leq \frac{\sqrt{2n} + 1}{\sqrt n - 1}$$
and
$$a_n = \frac{\...
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0
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Computation of limit involving specific differences of logarithms
Hint
$$\sum_{k=a}^b \log(k)=\log \Bigg(\prod_{k=a}^b k\Bigg)=\log \Bigg(\prod_{k=1}^b k\Bigg)-\log \Bigg(\prod_{k=1}^a k\Bigg)= ???$$
- 256k
0
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Does this prove that the factorial grows faster than the exponential?
Yes, your method shows specifically that $n!=\omega(e^n)$, since $L_n\to\infty$. There is nothing special about $e$ here, so you can use the same idea to show that $n!=\omega(\alpha^n)$ for any $\...
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1
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How to prove $\int_{\mathbb R} \nabla_u F (x, u_k(x))\cdot\varphi dx \to \int_{\mathbb R} \nabla_u F (x, u(x))\cdot\varphi dx$ for all test function?
Let $K$ be the compact support of $\varphi$. Then we can write:
$$
\int_{\mathbb{R}} \nabla_u F(x, u_k(x))\cdot \varphi ~\mathrm{d}x = \int_K\nabla_u F(x, u_k(x))\cdot \varphi ~\mathrm{d}x
$$
Because ...
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3
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Iterating $\log(x\log(x\log(...)))$
For $x>e,$ $\ln(x)>1$ then let's define the sequence $f_0(x)=x,$ $f_{n+1}(x)=\ln(xf_{n}(x)) $
to prove the limit $\displaystyle \lim_{n\to\infty} f_n(x)$ always exist for any $x>e$ I will use ...
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4
votes
Does this prove that the factorial grows faster than the exponential?
$$
\frac{n!}{e^n} = \prod_{i=1}^n\frac{i}{e}.
$$
Since
$$
\lim_{i \to \infty} \frac{i}{e} = \infty,
$$
clearly the factorial is growing faster than the exponential.
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1
vote
Accepted
Does $ \lim_{x \to 0^+} (1 + x)^{\ln x}$ exist?
Suppose there's a limit, say
$$
L = \lim_{x \to 0^+} (1 + x)^{\ln x}.
$$
Apply the natural logarithm, and note that $u \mapsto \ln u$ is continuous, so we can pass the limit through the logarithm:
\...
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3
votes
Accepted
Can the following $\infty - \infty$ limit be solved without using l'Hôpital rule?
$\lim\limits_{x\to0}\dfrac{\sin^2x-x^2}{x^2\sin^2x}=\lim\limits_{x\to0}\dfrac{(\sin x-x)(\sin x+x)}{x^2\sin^2x}$
$=\lim\limits_{x\to0}\dfrac{(-\frac16x^3\cdots)(2x \cdots)}{x^4}\lim\limits_{x\to0}\...
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4
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Accepted
What is $\lim_{x\to\infty}\frac{\int_{0}^{x}\cos\{t-\cos t\}dt}{x}$?
The limit is
$$\sin(1) \approx 0.8414709848.$$
As others have said, invalid approximations using floating point arithmetic are likely to blame for the values in the table with $x > 10^{16}$.
Let $f(...
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1
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Limits involving points at infinity in the extended complex plane
The simplest way to justify the point at infinity in (complex) projective geometry is in terms of homogeneous coordinates. Thus, the ordinary points $z\in \mathbb C$ correspond to the pair $[z,1]$ in ...
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1
vote
Accepted
Finding the points of discontinuity of $f(x)= [x]\sin\left(\frac{π}{[x+1]}\right)$
A possibile path
Let
$$g(x) = x\sin\left(\frac{\pi}{x+1}\right).$$
You get
$$g'(x) = \sin\left(\frac{\pi}{x+1}\right)-\frac{\pi x}{(x+1)^2}\cos\left(\frac{\pi}{x+1}\right).\tag{1}\label{1}$$
Now, for ...
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2
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Limits involving points at infinity in the extended complex plane
Definition of complex infinity
You have the complex plane $\mathbb C$. Then, we add an arbitrary point $\infty$ to $\mathbb C$ and say that if you go away from the origin (from any direction or path) ...
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