Given: $f(x)$ is defined on $\mathbb{R}$ and $|f(x) -f(y)| \le |x-y|^\alpha$. Which of the following statements are true? I. If $\alpha > 1$, then $f(x)$ is constant. II. If $\alpha = 1$, then $f(x)$ is differentiable. III. $0 < \alpha < 1$, then $f(x)$ is continuous. Answer: I $-$ true, II $-$ false, III $-$ […]

I’m working through some analysis books, and while working through the section on directional derivatives, I searched here and found this answer, which states Let $f: R^2 \to R$ be defined by $f(x,y)= \frac{x^3y}{x^4+y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0) = 0$. This function is continuous; its directional derivative is defined at each point in […]

Let $f(x)=x^2\sin(1/x)$ for $x≠ 0$ and $f(0)=0$ for $x=0$. Is $f'$ continuous at $0$? My attempt: $f'(x)=2x\sin(1/x)-\cos(1/x)$. Since when $x$ goes to $0$, the limit of $\cos(1/x)$ does not exist, it is not continuous. But I’m not sure since we did define $f(0)=0$…

Calculate the surface area of the ellipsoid that is given by rotating $\frac{x^2}{2}+y^2=1$ around the x-axis. My idea is that if $f(x)=\sqrt{1-\frac{x^2}{2}}$ rotates around the x-axis we will end up with the same figure. The formula for calculating the surface area is: $ 2\pi \int_{-\sqrt{2}}^{\sqrt{2}} f(x)\sqrt{1+f'(x)^2}dx $ and with $f'(x)^2 = \frac{x^2}{4-2x^2}$ the integral becomes […]

The textbook explanation shows that L’Hôpital’s rule can be used on a rational function ${f(x)}\over {g(x)}$ if it is continuous, and $\lim_{x \to c}f(x) = \lim_{x \to c}g(x) = 0$ or $\pm \infty$, and $g'(x) \neq 0$. The textbook doesn’t explain, however, what steps are necessary to take before I dive blindly into applying L’Hôpital’s […]

We say that a continuous function $f:\mathbb{R}\to\mathbb{C}$ is almost periodic in the sense of Bohr if: For every sequence $(t’_n)_{n\geq0}$, there’s a sub-sequence $(t_n)_{n\geq0}$ such that $f(t+t_n)$ converges uniformly in $\mathbb{R}$ to a function $g(t)$ i.e. $$\sup_{t\in \mathbb{R}}|f(t+t_n)-g(t)|\to 0, \ \ when \ \ n\to +\infty.$$ This class of functions was proved to be the […]

I need somebody to explain me why $$\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$$ I don’t get it. I’d say the left hand side is 0, because there is no change in $y^2$ when x changes slightly.

Need some help finding this limit: $$\lim_{x\rightarrow1}\frac{\frac{1}{\sqrt{x}}-1}{x-1}$$ Here is what I have so far: $$\lim_{x\rightarrow1}\dfrac{\dfrac{1-\sqrt{x}}{\sqrt{x}}}{x-1}$$ $$\lim_{x\rightarrow1}\dfrac{1-\sqrt{x}}{\sqrt{x}}\cdot\dfrac{1}{x-1}$$ $$\lim_{x\rightarrow1}\dfrac{1-\sqrt{x}}{x\sqrt{x}-\sqrt{x}}$$ At this point I keep getting results I don’t like, I have tried multiplying by the conjugate but I keep getting denominators of $0$. What am I missing here? Thanks

prove that the rose $r=\cos(n\theta)$ (in the polar plane) has $2n$ “petals” when $n$ is even. How can I start this demonstration? I would appreciate your help

Given a succession of real numbers $\{a_n\}_n$ with $n \in \mathbb{N}$, I would like to prove, or find a counterexample, of the fact that if $|a_n|<1$ from a $n$ on then, then $\begin{aligned} \sum_{n=0}^{\infty} (a_n)^n \end{aligned}$ converges. Someone could give me a hand? Thank you!

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