# Tag Info

43

The rule $(fg)'=f'g+fg'$ works where $f$ and $g$ are differentiable. And $\sqrt[3]x$ is not differentiable at $x=0$.

41

Two functions are typically defined to be equal if and only if they... Share the same domain Share the same codomain Take on the same values for each input. Thus, functions $f,g : S \to T$ for sets $S,T$ have $f=g$ if and only if $f(x) = g(x)$ for all $x$ in $S$. For functions with holes, we typically restrict the domain by ensuring the values where the ...

41

When you have$$\lim_{n\to\infty}\frac{\ln(n)}{\ln(4n)},$$the denominator doesn't increase at a much faster rate than the numerator. As a matter of fact, since we have$$(\forall n\in\mathbb N):\ln(4n)=\ln(4)+\ln(n),$$they increase at the same rate. And now it is easy to see that the limit is indeed $1$.

38

WolframAlpha understands the expression $\sqrt[3]{x}$ for negative x in a different way than you expect. Try this: lim\frac{\sqrt{1-x}-3}{2+surd(x,3)} as x to -8

34

Note that for any $x$ we have $x\cdot 0=0$ and therefore $$\lim_{x\to\infty} (x\cdot 0) =\lim_{x\to\infty} 0=0$$

33

Table of Content. Heuristic argument Elementary proof, version 1. Elementary proof, version 2. (NEW!) 1. Heuristic argument. Although far from being rigorous, one can provide a heuristic computation which explains why we expect the answer to be $\frac{1}{2}$. Notice that $$\frac{n^{n+j}/(n+j)!}{n^n / n!} = \begin{cases} \prod_{k=1}^{j} \frac{n}{n+k}, &... 28 This is just another way of saying what the others told you.$$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3} \ne \lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}$$The theorem is IF \displaystyle \lim_{x\to 0}f(x) = L and \displaystyle \lim_{x\to 0}g(x)=M, where M, N \in \mathbb R, THEN \displaystyle \lim_{x\to 0}(f(x)-g(x))=L-M ... 26 Outline: Use the inverse function of y=x-\sin x to express f_\infty(x). Use integral of inverse functions and dominated convergence theorem to prove L=2. Claim:$$L=2.$$Proof: Obviously y=t-\sin t is injective on t\in[0,\pi]. Define y=\operatorname{Sa}(t) as the inverse function of y=t-\sin t on t\in[0,\pi]. Therefore,$$t-\sin t =x \...

26

As others have said, $\lim_{x\to \infty} 0 \times x = \lim_{x\to\infty} 0 = 0$. I'm going to expand a bit more on "$0 \times \infty$ is undefined". We can't do operations with $\infty$ directly, as you know. But we can do operations with "functions with limit $\infty$", and if they behave well enough then that might give us reasonable definitions of things ...

26

If I were taking that exam, I'd speculate covergence and write the integrand for $A_\infty$ as $$S_\infty(x) = \frac{1}{ 1 + \frac{1}{1+\frac{1}{1+\frac{1}{\ddots} }}} = \frac{1}{1+S_\infty(x)}$$ Solve the resulting quadratic for $S_\infty^2(x) + S_\infty(x) -1 = 0$ for $S_\infty(x)=\frac{-1+\sqrt{5}}{2}$. Then we immediately have $A_\infty = S_\infty$. ...

25

In general, the statement $$(AB)^n=A^nB^n$$ is false for square matrices. So it's not true in general that, from $A=PDP^{-1}$ it follows that $A^n=P^nD^n(P^{-1})^n$. Rather you should note that $$A^2=(PDP^{-1})(PDP^{-1})=PDP^{-1}PDP^{-1}=PDDP^{-1}=PD^2P^{-1}$$ and, by easy induction, $$A^n=PD^nP^{-1}$$ for every $n$. Do you see the difference? Now, ...

23

Mathematicians (but not all calculus books) mean "continuous at every point of its domain" when they say a function is "continuous." The functions $f(x) = 1/x$ and $f(x)=\ln x$ are continuous functions.

22

The product rule trick still works. If $\lim_{x \to 0} f(x)/x = R \in \mathbb R$, and obviously $\lim_{x \to 0} x = 0$, it follows that $$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{f(x)}{x} \times x = R \times 0 = 0.$$

22

\begin{align} \lim_{n\to\infty}n\int_0^1\frac{x^n}{1+x+x^n}\,\mathrm{d}x &=\lim_{n\to\infty}\int_0^1\frac{x^{1/n}}{1+x^{1/n}+x}\,\mathrm{d}x\tag1\\ &=\int_0^1\frac1{2+x}\,\mathrm{d}x\tag2\\[3pt] &=\log\left(\frac32\right)\tag3 \end{align} Explanation: $(1)$: substitute $x\mapsto x^{1/n}$ $(2)$: Dominated Convergence $(3)$: evaluate ...

20

Hint. Note that $$\Biggl({x+3\over x+8}\Biggl)^x=\frac{(1+\frac{3}{x})^x}{(1+\frac{8}{x})^x}.$$ Moreover, for $a\not=0$, after letting $t=x/a$ we have that $$\lim_{x\to \infty}(1+\frac{a}{x})^x=\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{ta}=\left(\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^t\right)^a=e^a.$$ where we used the limit which defines the ...

19

Your quote isn't a definition of a limit, but an English language description of what it computes. If you look at the actual definition, such as the usual "epsilon-delta" definition, you'll see that it handles constant functions just fine, and in fact you have $$\lim_{x \to a} c = c$$ So, what has happened here is that there is a miscommunication —...

19

$G$ is continuous on the domain $[0,3)\cup(3,6]$. Referring to the aforementioned definition (1) that the limits converge to the actual value at this point. 3 is not in the domain. For every point in the domain of $g$, we have the required convergence.

19

You can just squeeze it: $$1\leq(n!)^{1/n^2}\leq (n^n)^{1/n^2}=n^{1/n}\to 1.$$ So $\lim_{n\to\infty} (n!)^{1/n^2}=1$.

18

The logics for the ratio of polynomials: From a ratio with equal degrees such as $$\lim_{x\to\infty}\frac{3x^2+2x+1}{2x^2+7x-4}$$ we can rewrite $$\lim_{x\to\infty}\frac{3+\dfrac2x+\dfrac1{x^2}}{2+\dfrac7x-\dfrac4{x^2}}.$$ In the latter expression, it is clear that the terms with a denominator will vanish and all that remains is $\dfrac32$. If the ...

18

Similar to $\int_0^\pi\sin(x+\sin(x+\sin(x+\cdots)))\,\mathrm dx=2$, we only need to concern the integral on $[0, \pi/2]$ as mentioned by @Stijn Dietz. Let $\operatorname{Sb}(x)$ be the inverse function of $x\sin x$ on $[0, \pi/2]$ (such function exists by the injectivity). Therefore, $$t\sin t =x\implies t=\operatorname{Sb}(x).$$ Assume $f_\infty$ exists ...

18

The answer to this question depends on the irrationality measure $\mu(\pi)$ of $\pi$, in a way which means it is unsolved. (The current state of the art is that $2 \leq \mu(\pi) \leq C$, where $C \approx 7.6$.) Suppose that $\mu(\pi)>4$. Then there exist infinitely many pairs of integers $(p,q)$ such that $$\left|\pi - \frac{p}{q}\right|<\frac{1}{q^4}... 18 As requested by the OP in a comment deleted by a moderator, switching of limit and integral sign is avoided as it requires a higher-than-expected level of knowledge for justification. Thus, a ‘simpler’ approach is presented. By the substitution t=\sqrt x,$$A_n=\int^1_0f_n(t^2)(2tdt)$$f_n(t^2) is of the form$$f_n(t^2)=\frac{a_n+b_nt}{c_n+d_nt}$$We ... 18 Why not apply the MVT? The integral is$$F(x+1)-F(x)=\frac{F(x+1)-F(x)}{x+1-x}=F'(x_0)$$for an x_0 between x and x+1. Now F'(x) certainly converges to 1 if x tends to infinity. 17 It seems there is some confusion between sets and sequences. In this example, what you have written as "{1, 2, 1, 4, 1, 6, ...}" is not meant to be a set but rather a sequence (a_n) with a_0=1,a_1=2,a_2=1,a_3=4,\dots. In particular, then, when we say a point x is a limit point of (a_n), this means that for every neighborhood U of x, there exist ... 17 The proof that the product of continuous functions is also continuous does not need to use square roots. You could add the condition that 0 < \epsilon < 1, in which case \epsilon^2 < \epsilon. Or you could use a convergent sequence x_n such that \displaystyle \lim_{n \mathop \to \infty} x_n = a, and show that$$\lim_{n \to \infty}f(x_n)g(...

17

$$\lim_{x \to 1} \dfrac{x^{2019}-x^{1875}}{x-1} = \lim_{x \to 1} \left( x^{1875}\cdot \dfrac{x^{144}-1}{x-1}\right)$$ The product of the limits is equal to the limit of the product so long as both limits exist.

16

It is a quite common notation, if used, for multiplication, i.e. $$5.3=5\cdot3=5\times3=15$$ In your case $$dy/dx.dx/dt=\frac{dy}{dx}\times\frac{dx}{dt}$$ and $$e^x=1+x+\frac{x^2}{1.2}+\frac{x^3}{1.2.3}+\cdots=1+x+\frac{x^2}{1\times2}+\frac{x^3}{1\times2\times3}+\cdots$$

15

Unlike in the case of iteration $t \mapsto \sin(x+t)$, $f_n(x)$ does not seem converge beyond a certain threshold of $x$. Indeed, plotting the graph of $f_n)$ on $[1,\pi]$ and $201 \leq n \leq 264$ gives which clearly demonstrates the chaotic behavior as in the logistic map. This can also be glimpsed by the fact that the iteration $t \mapsto \sin(xt)$ ...

15

L'Hospital's rule contains an assumption that $\lim_{x \to a} f'(x)/g'(x)$ exists, which is not true in this case.

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