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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 $\lim x_n=1$.


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 ...


7

Use the fact that $x^2 \leq x^2 + |y|$ and then apply the squeeze theorem.


5

Every point on the limiting curve has a distance 1 to the $x$-axis along the tangent line at that point. Intuitively, this should be plausible. More formally, let $x=f(y)$ be the equation for the curve. The equation for the tangent line at $(f(y_0),y_0)$ is $$x-f(y_0)=f'(y_0)(y-y_0).$$ This meets the $x$-axis at $(-y_0f'(y_0)+f(y_0),0)$. The condition ...


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 $\lim x_n=1$.


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 $\left(\frac{\sin x}{x}\right)^{-2}$ ...


5

This is the same as the following argument $$\lim_{n\to\infty}\frac1n=\lim_{n\to\infty}\frac2n=0$$ $$\implies\frac1n=\frac2n$$ $$\implies1=2$$ Why would the first equality imply the second?


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 ...


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.


4

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 ...


4

There are several ways to proof this. It depends on what you already know. If you know that $\lim_{n\to\infty} \frac{1}{n}=0=\lim_{n\to\infty}-\frac1n$ then since $-\frac{1}{n}\leq \frac{(-1)^n}{n}\leq \frac{1}{n}$ It is $\lim_{n\to\infty} \frac{(-1)^n}{n}=0$ by the squeezing theorem. Or simply since the sequence $a_n=\frac1n$ converges to $0$ and $(-1)...


4

You can't, in general, split up an expression and take limits of different parts separately. That's why you don't get the right result the first time.


3

Note that as $x\to +\infty$, $$\cot^{-1}(x)=\arctan(1/x)=\frac{1}{x}-\frac{1}{3x^3}+o(1/x^3).$$ Hence, from your work, $$\frac{-x^3}{2}\bigg(\frac{-x}{1+x^2} + \cot^{-1}(x)\bigg)= \frac{-x^3}{2}\bigg(-\frac{1}{x}\frac{1}{1+\frac{1}{x^2}} + \frac{1}{x}-\frac{1}{3x^3}+o(1/x^3)\bigg)\\ =\frac{-x^3}{2}\bigg(-\frac{1}{x}+\frac{1}{x^3} + \frac{1}{x}-\frac{1}{3x^3}...


3

One can also express the area (as noted in the question) as a sum of trapezoids, obtained by dropping perpendicular lines from $E$, $G$, $I$, $K$, ... to the $x$-axis. The vertical bases of $k$-th trapezoid measure $r^k$ and $r^{k+1}$, while its height is $(1-r)\sqrt{1-r^{2k}}$. Hence total area $S$ can also be computed from: $$ S=\sum_{k=1}^\infty{1\over2}(...


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$


3

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 $$\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}} =\frac{x+2-2-\frac x2+O(x^2)}{3x-1+1+\frac x2+O(x^2)} \\=\frac{\frac x2+O(x^2)}{\frac72x+O(x^2)}=\frac17+O(x)\underset{x\to0}\longrightarrow\frac17 $$


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 $x_n \to 1$.


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 ...


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 $$0\leq \frac{|x|}{\sqrt{x^{4}+4x^{2}+7}}\leq |x|.$$


3

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


3

Hint. After reading your work, you may show that $$\frac{x^2y}{x^2+|y|}=\frac{\rho^2\cos^2(\theta)|\sin(\theta)|}{\rho\cos^2(\theta)+|\sin(\theta)|} \leq \rho^2$$ that is $$\rho^2\cos^2(\theta)|\sin(\theta)|\leq \rho^3\cos^2(\theta)+\rho^2|\sin(\theta)|.$$


3

Hint: since $0\le|\sin(\theta)|,|\cos(\theta)|\le 1$ you have that $$0\le\left|\frac{\rho^2\cos^3(\theta)\sin(\theta)}{\rho\cos^2(\theta)+|\sin(\theta)|}\right| \le \frac{\rho^2}{\rho[1-\sin^2(\theta)]+|\sin(\theta)|}\le \begin{cases} \rho^2 & 1<\rho\\ \\ \rho & 0< \rho\le 1 \end{cases}$$ since $0\le 1-\sin^2(\theta)\le 1$ and $|\sin(\theta)|\...


3

Without De l'Hopital, you could simply go this way (recalling the factorization $a^3-b^3 = (a-b)(a^2+ab+b^2))$. $$\lim_{x\to 0}\frac{\sqrt[3]{ax+b}-2}{x}=\lim_{x\to 0}\frac{ax+b-8}{x\left(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\right)}.$$ Again here, as in comments, we need the numerator to be $0$ in $x=0$, which means $b=8$. Therefore the limit becomes $$\...


2

We have that $$\text{arccot}{(x)}=\arctan{\left(\frac1x\right)}=\frac1x-\frac1{3x^3}+o\left(\frac1{x^3}\right)$$ as $x\to\infty$. Also note that as $y\to0$ we have $$\ln{(1+y)}=y+o(y)$$ Hence our limit is simply $$\begin{align} \lim_{x\to\infty}x^2\ln{(x\text{arccot}{(x)})} &=\lim_{x\to\infty}x^2\ln{(x(1/x-1/(3x^3)+o(1/x^3)))}\\ &=\lim_{x\to\infty}x^...


2

All that you have proved was that if your limit exists, then it cannot be greater than $0$. You can prove that it is $0$ using the fact that$$\left\lvert(x^2-y^2)\sin\left(\frac1{x^2+y^2}\right)\right\rvert\leqslant x^2+y^2.$$


2

$$ \lim_{x\to 1}\frac{P(x)}{(x-1)^2}\hbox{ exists}\implies\lim_{x\to 1}P(x) = 0\implies P(1) = 0\implies P(x) = (x - 1)Q(x)\hbox{ with $Q(x)$...} $$ Can you say why all the $\implies$ are true? Can you continue?


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 ...


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 }\...


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 $n^{1/n} \to 1$, $L = 1$.


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 $L$ ...


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