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if a graph has vertical asymptote, the derivative must also have a vertical asymptote too, right? No. A counterexample: $$f(x)=\frac{1}{x}+\sin\left(\frac{1}{x}\right)$$ This function is monotone and has a vertical asymptote at $x=0$. But its derivative has no limit.

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Not quite. Suppose $y = f(x)$ has a vertical asymptote at $x=a$ in the sense that $\lim_{x \to a\pm} f(x) = +\infty$ or $-\infty$. Then $\lim_{x \to a\pm} f'(x)$ can't be a finite number. It could be $+\infty$ or $-\infty$, but it also might not exist at all. In fact, $f'$ might not exist at all. In the other direction, $f'$ can have a vertical ...

16

In short: no, you did not do it correctly. The reason is that $\infty\cdot 0$ is one of the indeterminate forms, you cannot conclude it's equal to zero. The rationale being: it can be anything. For instance, \begin{align} \underbrace{n}_{\to \infty}\cdot \underbrace{\frac{1}{n}}_{\to 0} &\xrightarrow[n\to\infty]{} 1\\ \underbrace{n^2}_{\to \infty}\... 8n-n\sqrt{1-\frac{5}{n}}=\frac{(n-n\sqrt{1-\frac{5}{n}})(n+n\sqrt{1-\frac{5}{n}})}{n+n\sqrt{1-\frac{5}{n}}}=\frac{5}{1+\sqrt{1-\frac{5}{n}}}\rightarrow \frac{5}{2}, \space \text{as} \space \space n \rightarrow \infty.$$8 The first branch point is associated with f, and happens at x=0. We note that:$$g\circ f(x) = \begin{cases} g(2x+1), & \text{if $x≤ 0$} \\ g(x^2), & \text{if $x$ > 0} \end{cases}$$Let's just consider each case separately. First, g(2x+1) when x≤0. We remark that x≤0\implies 2x≤0\implies 2x+1≤1 so we see that x≤0\implies g(2x+1)=-... 8 Expand the cube and divide top and bottom by n to get$$\begin{align*}\lim_{n \to \infty} \frac{n\sqrt{n} +3n + 5\sqrt{n} + 2}{n + \sin n} &= \lim_{n \to \infty} \frac{\sqrt{n} + 3 + \frac{5}{\sqrt{n}} + \frac{2}{n}}{1 + \frac{\sin n}{n}} \end{align*}$$from which you can see the numerator diverges off to \infty whilst the denominator tends to 1. ... 8 Taking @SiongthyeGoh's comment, in the above picture, AE=1, \angle BAE=x, \angle BAC=\beta, and \angle BAD=\alpha. It follows that DE=\sin(x+\alpha) and CE=\sin(x+\beta). Applying the cosine theorem to \triangle DCE, the given expression is equal to DC^2, hence is a constant. 7 You might represent this in a nicer way by taking u = x+t, v = x-t, so x = (u+v)/2 and t = (u-v)/2, making the equation$$ f(u) - f(v) = u^2 - v^2 $$Next notice you can separate the u's and the v's. What does that tell you? 7 The notation f:B\to A is typically used to denote that f is a function from B into A. Thus, saying$$A^B = \{f :B \to A| f \text{ is a function}\}$$is like saying A^B is the set of all functions f:B\to A such that f is a function. So there is some redundancy. I, personally, would go with the first option, however your intentions seem ... 7 It's sometimes a good idea to turn a sum of sines and cosines into a single trigonometric term when solving equations of the form a \sin x + b\cos x = c, in this case:$$\sqrt{3}\cos v - \sin v \equiv 2\cos \left(v + \frac{\pi}{6}\right)$$So, if you set x = v + \frac{\pi}{6} you need only solve \cos x = \frac{1}{\sqrt{2}} which you can then do, I'm ... 6 Put x=t and the solution drops out. 6 That's a fairly standard rewrite. If you need to find something like$$ \lim_{x\to a} f(x)g(x) $$where \lim_{x\to a}f(x)=0 and \lim_{x\to a}g(x)=\infty, you can define h(x)=\frac{1}{g(x)} and consider \lim_{x\to a} \frac{f(x)}{h(x)}. You should be able to use the form of l'Hôpital you know on that. Of course that requires g to be well-behaved ... 6 l'Hopital's rule only works for the limits "0/0" and "\infty/\infty". To compute this limit try this:$$\frac{\sin x+\cos x}x=\frac{\sin x}x+\frac{\cos x}x$$5$$g(x) = \begin{cases} -x, & x < 2 \\ 5, & x \ge 2 \end{cases} $$Therefore$$g(f(x)) = \begin{cases} -f(x), & f(x) < 2 \\ 5, & f(x) \ge 2 \end{cases} $$So now we need to know when f(x) < 2 and when f(x) \ge 2.$$f(x) = \begin{cases} 2x + 1, & x \le 0 \\ x^2, & x > 0 \end{cases} $$Let's look at one piece of f ... 5 No, you can't do it like that: when you have a product where one factor has limit \infty and the other has limit 0, you cannot apply a theorem on products of limits. The theorem helps when both limits are finite and you just multiply them; it also works when one limit is \infty (or -\infty) and the other one is either infinite or *finite and not 0”... 5 I'll remark on two of your four problems, and hopefully that will be enough to get you going on the others. First, you have already been given the domain and codomain for your function f, namely \mathbb{N} and \mathbb{N}. Thus, for every one of of your problems you are considering the mapping f\colon\mathbb{N}\to\mathbb{N} defined by ___ ; that is, ... 5 Rewriting would give f(x)+f(\frac{1}{x})=\dfrac{x^2-12x+1}{2x}=\frac{1}{2}(x+\frac{1}{x})-6 Then, it is easy to see a possible function is f(x)=\frac{1}{4}(x+\frac{1}{x})-3 So, if you put \frac{1}{x} in the argument, you would get the same function back. However, as mentioned in comments, f(x)=\frac{x}{2}-3 is also a valid solution. Therefore , ... 4 No. There is an infinite-dimensional space of homeomorphisms of the circle; the space of Mobius transformations is the 3-dimensional PSL_2(\Bbb R). But more pointedly, the restriction of a Mobius transformation to the ideal boundary can have at most two fixed points if it's not the identity; but any closed subset of S^1 is the set of fixed points of some ... 4 I don't think Robert Israel's answer quite corresponds to the question as now phrased, and at the moment I don't have an answer that precisely does so either. However, I might let X be a random variable that has a continuous uniform distribution on the interval [0,1] (so that its probability of falling into any subinterval is the same for all ... 4 Let$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\color {blue}{2\cos(\alpha-\beta)\sin(x+\alpha)}\sin(x+\beta)$$Now use 2\cos A \sin B = \sin (A+B) - \sin (A-B).$$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)-\sin\left(\alpha-\beta -x-\alpha\right)\right]\sin(x+\beta)E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(...

3

The brakes reduce it's velocity by $10$ m/s in every second. So the acceleration is $-10 \, \text{m/s}^2$. So $$y(t)=v-10\,\text{ms}^{-2}t$$ $$x(t)=100\,\text{ms}^{-1}t+\frac{1}{2}(-10\,\text{ms}^{-2})t^2$$

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If $g(x)$ is a differentiable function, the derivative of $g(x)e^{-x}$ is given by $\left(g'(x)-g(x)\right)e^{-x}$. If $g(x)=\sum_{k=0}^{2n}\frac{x^{k}}{k!}$ we have that $g'(x)-g(x)$ is a monomial and an even function, hence $f(x)=g(x)e^{-x}$ has a unique stationary point at the origin and is a decreasing function. Since $\lim_{x\to +\infty}g(x)e^{-x}=0$ ...

3

You're making this much too complicated. If such a $\psi$ exists, to show that $f$ is surjective, you just need to show that for every $b \in B$, there exists $a \in A$ such that $f(a)=b.$ You can define such an $a$ explicitly using $\psi$ and $b.$ For the implication from left to right, note that you'll need to use the axiom of choice. In fact, that ...

3

HINT Use $\sin^2t=\frac {1-\cos2t} 2$ and $\cos2p+\cos2q=2\cos(p+q)\cos(p-q)$ formulas

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a solution is $$g(x) = \sum_{n=-\infty, n \ne 0}^\infty e^{-n^4 (x-n)^2}$$ it is real analytic since it is complex analytic it is $L^1$ since $$\|g\|_{L^1} = \sum_{n=-\infty}^\infty \int_{-\infty}^\infty e^{-n^4 (x-n)^2} dx = \sum_{n=-\infty, n \ne 0}^\infty \frac{1}{n^2}\int_{-\infty}^\infty e^{-x^2} dx = 2\frac{\pi^2}{6}\sqrt{2\pi}$$ and clearly $g(x) \... 3 Let$x^4+5=z$, then you got$dz=4x^3dx$, so $$\int 4x^3(x^4+5)^5dx=\int z^5dz=\frac{z^6}{6}+K=\frac{(x^4+5)^6}{6}+K$$ 3 The simple answer, in the spirit of the comments, is that all polynomials are functions but not all functions are polynomials. A function is simply a rule that assigns a value in the codomain to every value in the domain. A couple simple examples of functions that are not polynomials are$\sin x$and$|x|$. A less simple one is the Dirichlet function which ... 3 A relationship from X to Y where every element of X is mapped to an element of Y is called a "Total function". Relationships where some elements of X are unmapped are called "Partial functions". However formally all functions are total and people say "Total function" only where it would be ambiguous. 3 Divide the equation by$\;2\;$: $$\frac1{\sqrt2}=-\frac12\sin x+\frac{\sqrt3}2\cos x=\sin\frac\pi3\cos x-\cos\frac\pi3\sin x=\sin\left(\frac\pi3-x\right)\implies$$ $$\frac\pi3-x=\begin{cases}\cfrac\pi4\\{}\\\cfrac{3\pi}4\end{cases}\implies \ldots$$ Observe this is similar to the other answer but, perhaps, a little, very little, easier to understand. 3 The limits for$+3$and$-3$have nothing to do with each other. The notation you want for limits is$3^+$and$3^-$or$-3^+$and$-3^-$You are a little confused by the notation. Let's say we want to find the limit of the function for$x \to 3$. Then we need to find the limit for$x \to 3^+$which means$x$approaches$3\$ from the right side. Then we need ...

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