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7

It is true that $\ln(x^2-1)$ is an antiderivative of $\dfrac{2x}{x^2-1}$ on an interval where $\ln(x^2-1)$ is defined and differentiable. But unless you want to get into complex values for logarithms of negative numbers, you won't want to use this for $-1 < x < 1$. Instead, for $-1 < x < 1$ you could take the antiderivative as $\ln(1-x^2)$. ...


4

Using polar: $\lvert\dfrac{r^3\cos^2\theta\sin\theta}{r^2(\cos^2\theta+r^2\sin^2\theta)}\rvert=\lvert\dfrac{r\cos^2\theta \sin\theta}{\cos^2\theta +r^2\sin^2\theta}\rvert\le\lvert\dfrac {r\cos^2\theta \sin\theta}{\cos^2\theta}\rvert=\lvert r\sin\theta\rvert\to0$, if $\theta\neq\dfrac{k\pi}2$. But it's easy to see the limit is $0$ when $\theta =\dfrac {k\pi}...


4

No, as the sine function shows. It has no limit at $\pm\infty$.


4

No, it doesn't. For example, $\sin$ and $\cos$ are infinitely differentiable functions, but they have no limit at $\pm \infty$.


4

Write $x=y-1$ so $x!=\Gamma(y)$ and we want $\lim_{y\to0}\frac{\ln\Gamma(y)}{\Gamma(y)}$. For small $y$, $\Gamma(y)\sim\frac{1}{y}$, so the limit is $\lim_{z\to\infty}\frac{\ln z}{z}=0$.


3

Hint. We have that for $n\geq 2$, $${n\over 1+\sqrt n } \cdot \cos\left({ n\pi \over n+1 }\right)=-\sqrt{n}\cdot{1\over \frac{1}{\sqrt n}+1 } \cdot \cos\left({\pi \over n+1 }\right)\\\leq -\sqrt{n}\cdot{1\over \frac{1}{\sqrt 2}+1 } \cdot \cos\left({\pi \over 3 }\right).$$


3

If $(x,y) \neq (0,0)$, then we have \begin{align} \left| \dfrac{x^2y}{x^2 + y^4} \right| &= \left| \dfrac{x^2}{x^2 + y^4} \right| \cdot |y| \\ &\leq 1 \cdot |y| \\ &= |y| \end{align} From here it's easy to give an $\varepsilon$-$\delta$ argument for why the limit is $0$.


2

The map $x\mapsto\ln(x^2-1)$ is an antiderivative of $\dfrac{2x}{x^2-1}$. But if you replace $x$ by to concrete numbers and compute their difference, all you get as number, not an antiderivative of that function. Besides, $\ln(0)$ is undefined. However, it is true that$$\lim_{t\to1^-}\int_{-t}^t\frac{2x}{x^2-1}\,\mathrm dx=0.$$That is so because, for each $...


2

Mainly because of the probability interpretation. If $\psi$ is an eigenfunction of the Schrodinger equation, then the statistical interpretation says that $\int_a^b|\psi(x)|^2\,dx$ gives the probability of finding the particle between $a$ and $b$. By the rules of probability, we must have $\int_{-\infty}^{\infty}|\psi(x)|^2\,dx=1<\infty.$ This certainly ...


1

Let $f(x,y)={\large{\frac{x^2y}{x^2+y^4}}}$. Let $x^2+y^2=r^2$, with $0 < r \le 1$. If $x\ne 0$, then \begin{align*} |f(x,y)|&=\left|\frac{x^2y}{x^2+y^4}\right|\\[4pt] &\le\left|\frac{x^2y}{x^2+x^2y^4}\right|\;\;\;\;\;\text{[since $x^2\le r^2\le 1$]}\\[4pt] &=\left|\frac{y}{1+y^4}\right|\\[4pt] &\le |y|\\[4pt] &\le r\\[4pt] \end{...


1

Neither. Counter-examples: $f(x)=e^{-x^2},$ and $f(x)=x^2.$


1

Since $\lim_{x \to - 1} \log(x!) = \log( (-1)! ) = \infty$ and $\lim_{x \to - 1} x! = \tilde{\infty}$ and the functions are obviously continuous as composition of continuous functions we can apply the rule of L'Hôpital: By the chain rule we have \begin{align} \lim_{x \to - 1} \frac{\log(x!)}{x!} & = \lim_{x \to - 1} \frac{\frac{d}{dx} \log(\Gamma(x + 1))}...


1

We want to prove that $$\lim_{h\to 0} \frac{g(c)-g(c-h)}{h}=g'(c)$$ If we take $k:= -h$, then $k\to 0$ as $h\to 0$, and $$\lim_{h\to 0} \frac{g(c)-g(c-h)}{h}=\lim_{k\to 0} \frac{g(c)-g(c+k)}{-k}=\lim_{k\to 0} \frac{g(c+k)-g(c)}{k}=g'(c)$$ More general, first prove with the $\varepsilon,\delta$ definition that $$\lim_{x\to 0}f(x)=\lim_{x\to 0}f(-x)$$


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