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15

For a continuous function $f:[0,1]\to\mathbb{R}$, it is well-known that $f$ can be uniformly approximated by a linear combination of Bernstein polynomials, i.e., the function $$ F_n(x) = \sum_{k=0}^{n} \binom{n}{k}x^k(1-x)^{n-k}f\left(\frac{k}{n}\right) $$ converges uniformly to $f$ on $[0, 1]$ as $n\to\infty$. Your result is a special case with $x = \frac{...


10

You can write $P(x)$ as $(x-1)^2Q(x)+ax+b$, with $Q(x)\in\mathbb R[x]$ and $a,b\in\mathbb R$. So,$$\lim_{x\to1}\frac{P(x)}{(x-1)^2}=\lim_{x\to1}\left(Q(x)+\frac{ax+b}{(x-1)^2}\right).\tag1$$Since $\lim_{x\to1}Q(x)=Q(1)$, the limit $(1)$ exists if and only if the limit$$\lim_{x\to1}\frac{ax+b}{(x-1)^2}\tag2$$exists. But if the limit $(2)$ exists, then $a\...


9

Hint. If $n\geq 1001^2$ then $$0\leq \frac{n^{1000}}{n^{\sqrt{n}}}\leq \frac{n^{1000}}{n^{1001}}=\frac{1}{n}.$$


8

$$\frac{1}{\sqrt{2k}+\sqrt{2k-1}}=\frac{1}{\sqrt2\left(\sqrt{k}+\sqrt{k-\frac{1}{2}}\right)}<\frac{1}{\sqrt2\left(\sqrt{k}+\sqrt{k-1}\right)}=\frac{1}{\sqrt2}\left(\sqrt{k}-\sqrt{k-1}\right),$$ $$\frac{1}{\sqrt{2k}+\sqrt{2k-1}}=\frac{1}{\sqrt2\left(\sqrt{k}+\sqrt{k-\frac{1}{2}}\right)}>\frac{1}{\sqrt2\left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{1}{\sqrt2}\...


7

This is not true. Take $x_n=\frac 1 n$ and $x=0$ in the real line.


7

Let us ignore the $(-1)^n$ for the time being $$0<\dfrac{(n+1)!}{n^{n+1}}=\dfrac{n+1}n\cdot \prod_{r=2}^n\dfrac rn\cdot\dfrac1n<\dfrac{n+1}n\cdot\dfrac1n$$


7

No. Consider for instance the sequences $x_n=n+1$ and $y_n=n$, then $$\lim_n\frac{x_n}{y_n}=1,\quad\text{but}\quad \lim_n\frac{e^{x_n}}{e^{y_n}}=\lim_n e^{x_n-y_n}=e\not=1.$$ Another example: $x_n=n+\ln(n)$ and $y_n=n$, then $$\lim_n\frac{x_n}{y_n}=1,\quad\text{but}\quad \lim_n\frac{e^{x_n}}{e^{y_n}}=\lim_n e^{\ln(n)}=\lim_n n=+\infty.$$ As pointed out by ...


7

The denominator can be factored as $$(x^{1/3}-2)(x^{2/3}+2x^{1/3}+4)$$ so $$\lim_{x\to8}\frac{x^{1/3}-2}{x-8}=\lim_{x\to8}\frac1{x^{2/3}+2x^{1/3}+4}=\frac1{12}$$


6

First of all, if you do direct substitution, then yes, $\sec 0 = 1,$ but you still have to consider the arc sine function. And $\sin^{-1} (1) = \frac\pi2.$ So the result of direct substition by $x=0$ is $\sin^{-1} (\sec x) = \frac\pi2.$ But the direct substitution method agrees with the limit only if both of the following conditions are met: The limit ...


6

\begin{align} I&=\int_0^\infty\frac{x\ln x}{(1+x^2)^2}\ dx\overset{\large x\ \mapsto\frac1x}{=}-I\\ 2I&=0\\ I&=0 \end{align}


6

Your idea is correct, but you can be more concrete and more simple: It is $2^n=(1+1)^n=\sum_{k=0}^n \binom{n}{k}>\binom{n}{4}$ for $n$ big enough. $\binom{n}{4}=\frac{n!}{4!(n-4)!}=\frac{n(n-1)(n-2)(n-3)}{24}<\frac{n^4}{24}$ We can stipulate like this: $\lim_{n\to\infty} \frac{n^3}{2^n}\leq\lim_{n\to\infty} \frac{n^3}{n^4/24}=\lim_{n\to\infty} \...


6

When $x>0$, $|x|=x$, and so as $x\to\infty$ $\frac x{|x|}=\frac xx=1$. Sal is not substituting $\infty$ for $x$ here, which would be silly. Instead, the expression is just getting simplified to a constant, whose limit is that constant itself. And no, L'Hôpital's is unnecessary in this case.


6

Hint. $$ \frac{\sqrt[3]{x}-2}{x-8} = \frac{\sqrt[3]{x}-2}{(\sqrt[3]{x})^3-2^3} $$


6

Hint$$a^3-b^3=(a-b)(a^2+ab+b^2)$$


5

If $a_n>0$ and there exists limit $\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n}$, then $$ \lim_{n\rightarrow\infty} \sqrt[n]{a_n} = \lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n}$$ You can check that it's true as follows: If $\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = g > 0$ then for every $\epsilon>0$, $\epsilon <g$ there exists $N$ such ...


5

Note that $1+1/2+1/3+...+1/n \sim \gamma+\ln n$, where $\gamma=0.577\ldots$, is the Euler–Mascheroni constant. So your limit becomes $$ L =\lim_{n\rightarrow \infty} \frac{\ln n+\gamma}{\pi \ln n} = \lim_{n \rightarrow \infty} \left( \frac{1}{\pi}+ \frac{\gamma}{\pi\ln n} \right) =\frac{1}{\pi}. $$


5

Given $\varepsilon>0$, your formula $\delta:= \min\{1, \frac{\varepsilon}{9}\}$ works! Let's verify it. If $|x-1|<\delta$ with $\delta>0$, then $$|3x^2+1-4|=|3(x-1)(x-1+2)|\leq 3|x-1|(|x-1|+2)<3\delta(\delta+2).$$ Now, by the above definition, $\delta\leq 1$ AND $\delta\leq \frac{\varepsilon}{9}$ which implies $$3\delta(\delta+2)<3\cdot \...


5

Hint. Note that the sine values stays in the bounded set $[-1,1]$, and therefore for $(x,y)\not=(0,0)$, $$0\leq \left|(x^2+y^2) \sin\left( \frac{1}{x^2+y^2} \right)\right|\leq x^2+y^2.$$


5

It is natural to state that the domain $D$ of your function is the set of those real numbers $x$ at which that expression makes sense. So, in this case, $D=\mathbb{R}\setminus\{0,2\}$. And, for each $a\in D$, we have $\lim_{x\to a}f(x)=f(a)$ indeed. Therefore $f$ is continuous. What happens outside $D$ does not matter.


5

You are correct. The function is continuous, but the idea of continuity only applies at points in the domain. The points $0$ and $2$ are not in the domain. You are also right to be "worried" about those points. The function can be extended to a function $\hat{f}$ including $2$ in its domain (you must decide how to define $\hat{f}(2)$ in that case; the only ...


5

Let $t=x-\frac{\pi}{2}$, then we have to compute $$\lim_{t \to 0}\frac{1}{t}-\cot t.$$ Now use the series expansion for $\cot t$ given by $$\cot t=\frac{\cos t}{\sin t}=\left(1-\frac{t^2}{2}+o(t^3)\right)\frac{1}{t}\left(1+\frac{t^2}{6}+o(t^3)\right)=\frac{1}{t}-\frac{t}{3}+o(t).$$ So $$\lim_{t \to 0}\frac{1}{t}-\cot t=\lim_{t \to 0}\frac{t}{3}+o(t)=0.$$


5

For $t>1,$ $$\begin{align}(1+1/t)^{t}&= \exp{[t\ln (1+1/t)]}\\ & = \exp{[1-1/(2t) + O(1/t^2)]}\\ &= e\cdot(1+(-1/2t)+O(1/t^2)).\end{align}$$ Let $A= \int_0^1(1+1/t)^{t}\, dt.$ Then for $x>1,$ $$\begin{align}f(x)& = A + \int_1^x(1+1/t)^{t}\,dt\\&=A + e\int_1^x(1+(-1/2t)+O(1/t^2))\,dt\\ &= A + e(x-(\ln x)/2 +O(1)).\end{align}$$ ...


5

A more elementary solution : $$\lim_{x\to -\infty} \big( \sqrt{4x^2-x} +2x \big) =\lim_{x\to -\infty} \frac{-x}{\sqrt{4x^2-x} -2x} =\lim_{x\to \infty} \frac{x}{\sqrt{4x^2+x}+2x} =\lim_{x\to\infty} \frac{1}{\sqrt{4+\frac{1}{x}}+2} = \frac{1}{\sqrt{4}+2}=\frac{1}{4}$$ where I used that $$\lim_{x\to -\infty} f(x)=\lim_{x\to \infty} f(-x)$$


5

Choosing to limit $\delta$ is the right approach, but choosing to limit it to $|x - 2| < 1$ won't help you (and not for the reason you think). It's not a problem to get $-2 < x - 3 < 0$; all this means is that $|x - 3|$ will be between $0$ and $2$. The actual problem is that you're not bounding $|x - 3|$ away from $0$. You need $|x - 3|$ not just ...


5

Hint: $\displaystyle\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}=\frac1{\sqrt n}\times\frac1{\sqrt{1+\frac1n}+1}.$


5

Infinitely many continuous functions added together doesn't imply the result is also continuous. Take the Fourier series of a square wave function for example. All terms are continuous but the result is not. $$f(x)=\frac{4}{\pi}\sum_{n=1,3,5,...}{\frac{1}{n}\sin(\frac{n\pi x}{L})}$$


5

You want $$\lim_{k\to\infty}\frac{1}{k}\int_0^k e^{-y^2/2}dy=0$$by squeezing viz. $$0\le\frac{1}{k}\int_0^k e^{-y^2/2}dy\le\frac{1}{k}\int_0^\infty e^{-y^2/2}dy=\frac{\sqrt{\pi/2}}{k}.$$


5

You have : $$\lim\limits_{x\rightarrow +\infty}\sqrt{x-2}+\sqrt{x}=+\infty, $$so the limit of your expression is $0$.


5

Use the binomial theorem on $(1+x)^{1/3}$ to get $$ \sqrt[3]{n^3 + n^2 + 2n} = n\sqrt[3]{1 + n^{-1} + 2n^{-2}}= n\left[1 + \frac{n^{-1}+2n^{-2}}{3} - \frac{(n^{-1}+2n^{-2})^2}{9}+...\right] \\= n + \frac{1}{3} + \frac{5}{9n} + O(n^{-2}) $$ So $$ \lim_{n\rightarrow \infty}\sqrt[3]{n^3 + n^2 + 2n}-n = \frac{1}{3}. $$


5

Just as $\sqrt{1/x^6}$ goes to $0$, so does $\sqrt{x^6+4}$ go to $\infty$. You cannot substitute just one of these radicals and then simplify, and their unsimplified product is the indeterminate form $0\cdot\infty$ and so cannot be handled directly.


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