# Tag Info

1

here is a way to avoid l'hospitals but will use the fact $\lim_{n \to \infty} \dfrac{\ln n}{n} = 0$ let us do a change of variable $x = 1 - \dfrac{1}{n}.$ so $$\ln x \ln (1- x) = \ln(1 - 1/n) \ln(1/n)=\ln n(-\ln(1- 1/n)=\ln n\int_{1-1/n}^1\dfrac{dt}{t}$$ we can use the bound $\dfrac{1}{n} \le \int_{1-1/n}^1\dfrac{dt}{t} \le \dfrac{1}{n-1}$ to conclude $$... 1$$ \lim_{x\uparrow1} \ln x \ln(1-x) = \lim_{x\uparrow1} \frac{\ln x}{x-1}\cdot\frac{\ln(1-x)}{1/(1-x)}. $$Apply L'Hopital's rule to both fractions and you've got it. (In the first one, the numerator and denominator both approach 0; in the second, each approaches either +\infty or -\infty.) 1 With equivalents, it's very short: set x=1-u\enspace(u>0). Then \ln x\ln(1-x)=\ln(1-u)\ln u. As \ln(1-u)\sim_0 -u, we have \,\,\ln(1-u)\ln u\sim_0-u\ln u, which tends to 0 as u tends to 0. Hence$$\lim_{x\to 1_-}\ln x\,\ln(1-x)=\lim_{u\to 0_+}\ln(1-u)\ln u=0.$$1$$\frac{\log(1-x)}{\frac1{\log x}}\stackrel{l'H}\rightarrow\frac{-\frac1{1-x}}{-\frac{\frac1x}{\log^2x}}=\frac{x\log^2 x}{1-x}\stackrel{l'H}\rightarrow\frac{\log^2x+2\log x}{-1}\xrightarrow[x\to1^-]{}0 $$1 The proof is basically good but this part is a bit hard to follow and possibly misleading as regarding what the argument is. Taking \delta_{f\circ g}=\min\{\delta, \delta_1\}, we get that \forall x,y\in \Bbb{R} fulfilling |x-y|<\delta_{f\circ g}, |f(x)-f(y)|<\epsilon  and |g(x)-g(y)|<\delta. Also, you do not need the \min. The ... 1 It guarantees that (x-t)f''(t) is integrable, and that the fundamental theorem of calculus is applicable to it. Usually the statement of the fundamental theorem of calculus is limited to cases of a continuous integrand to avoid difficulties. EDIT: More precisely, what's important is that in the proof of the integration by parts formula, the FTC should be ... 1 You want f to have a continuous second derivative so that (x - t)f''(t) will be integrable over [a,x] for every x. Then E_1 will make sense. 0 You have the following ODE:$$g(x) + g'(x) = 12x^2 +2We can find the following integrating factor \begin{align} I &= e^{\int1dx} \\ &= e^{x}\end{align} Now multiplying through by I givese^xg(x) + e^xg'(x)=e^x(12x^2 +2)Thus we have that \begin{align}\frac{d}{dx}(e^xg(x)) &= e^x(12x^2 +2)\end{align} Integrating both sides w.r.t x gives ... 0 write your integral equations as(x+1)\int_0^x g dt - \int_0^x tg dt = x^4 + x^2$$differentiate once to get$$g(x) + \int_0^x g dt = 4x^3 + 2x \tag 1$$putting x = 0 gives g(0) = 0. differentiating (1) gives$$g^\prime + g = 12x^2 + 2 \tag 2$$try a particular to (2) solution to (2) in the form g = ax^2 + bx + c substituting in (2) we ... 0 you can supose your solution is of the form  ax^3+bx^2+cx+d+Ee^{-x}, you have term e^{-x} because homogenus equation g(x) + g'(x) =0 have this solution. 2 You have$$g(x) + g'(x) = 12x^2+2$$Most calculus text books cover basic ODE, the above is a first order linear ODE. To solve it, we introduce an integrating factor and notice$$ g(x) + g'(x) = 12x^2+2 \iff \frac{d}{dx} \left ( e^x g(x) \right ) =e^x( 12x^2 + 2) $$Now integrating the equation will give the result. 1$$\frac{n^2-n}{n^2+1}=1-\frac{n+1}{n^2+1}=1-\frac1{\frac{n^2+1}{n+1}}=1-\frac1{n-\frac{n-1}{n+1}}\implies\left(\frac{n^2-n}{n^2+1}\right)^{n+10}=\left(\frac{n^2-n}{n^2+1}\right)^{n-\frac{n-1}{n+1}}\cdot\left(\frac{n^2-n}{n^2+1}\right)^{10}\cdot\left(\frac{n^2-n}{n^2+1}\right)^{\frac{n-1}{n+1}}=$$... 2 Here is an alternative approach which only uses "elementary" calculations (i.e. no MacLaurin or any other kind of expansion, no continuity, no \log, no \exp): Note that$$\bigg(\frac{n^2-n}{n^2+1}\bigg)^{n+10}=\frac{\bigg(1-\frac{1}{n}\bigg)^n}{\bigg(1-\frac{1}{n^2}\bigg)^n}\cdot\frac{\bigg(1-\frac{1}{n}\bigg)^{10}}{\bigg(1-\frac{1}{n^2}\bigg)^{10}}$$... 1 I think it's simpler to rewrite it as an exponential and use the MacLaurin expansion of the logarithm:$$\begin{align} \lim_{n\to \infty }{\left({n^2-n\over n^2+1}\right)^{n+10}} &= \lim_{n\to \infty }{\exp\left((n + 10)\ln\left(1 - {\frac{n + 1}{n^2+1}}\right)\right)} =\\ &= \lim_{n \to \infty}\exp\left(-\frac{(n + 10)(n + 1)}{n^2 + 1} + o(1)\right) ...

1

Easier: write the original limit as $\bigg(\frac{1-\frac{1}{n}}{1+\frac{1}{n^2}} \bigg)^{n+10}$ and keep in mind $(1+\frac{1}{n^2})^n = e^{\frac{\log(1+\frac{1}{n^2}}{\frac{1}{n}}} \sim e^{\frac{1}{n}} = 1$. The last approximation uses Maclaurin expansion. Also keep in mind $(1+\frac{1}{n})^a \to_n 1$ if a is a constant.

2

Firstly, $$\int_0^1 f_n(x)dx=n\int_0^1 xe^{-nx^2}dx=n\left[-\frac{1}{2n}e^{-nx^2}\right]_0^1=n\left(\frac{1}{2n}-\frac{1}{2n}e^{-n}\right)=\frac{1}{2}-\frac{1}{2}e^{-n}$$ and thus $$\lim_{n\to\infty }\int_0^1 f_n(x)dx=\frac{1}{2}.$$ Secondly, $$\lim_{n\to\infty }nxe^{-nx^2}=0$$ and thus $$\int_0^1\lim_{n\to\infty }f_n(x)dx=0.$$ We conclude that ...

1

If $f(x,y)=|y- e^{-|x|}|e^{x^2}$, then for each fixed $y$ it clearly grows quickly with $x$. However, with $I=[0,1]$, we have $g(x)=0$.

0

Let $I=[\pi/2,3\pi/2]$. $$f(x,y):=\begin{cases}\sin(yx)x&\text{ for }y\in I\\\sin(\pi x/2)x&\text{ for }y<\pi/2\\\sin(3\pi x/2)x&\text{ for }y>3\pi/2\end{cases}$$

0

when you replace x you've to replace x everywhere. If $u=x+2$ then $x=u-2$ and $dx= du$. This means that your integral become $$\int (u-2) \sqrt u du = \int u \sqrt{u} du -2 \int \sqrt{u} du=\frac{2}{15}(3u-10) u^{\frac{3}{2}}$$ Now you go back by replacing $u$ by $x+2$. For the second integral please note that ...

2

the following expressions below are the same: $f^n(x)=(-1)^nn!x^{-(n+1)}$ and $f^{n-1}(x)=(-1)^{n-1}(n-1)!x^{-n}$ The only difference is that the first one starts at $n=0$ and the other one starts at $n=1$. You will yield the same answer no matter what formula you choose.

1

Hint for 1: If $u = x+2$ then $du = dx$ and $x = u-2$. Hence your first integral becomes $$\int x\sqrt{x+2}dx = \int(u-2)\sqrt{u}\space du$$ Remember this is single variable calculus, so it doesn't make sense to evaluate an integral like $\int x\sqrt{u}\space dx$ where both $x$ and $u$ are variables; the integral needs to be cast completely in terms of one ...

0

I think you can also divide top and bottom by x and get something like $\lim_{x \rightarrow 0^{+}} \frac{1-1}{x^{\frac{3}{2}}(1)}$

0

With MacLaurin formulas it's straightforward: $$\lim_{x\to 0^+}\frac1{\sqrt{x}}\left(\frac1{\sin x} - \frac1{x}\right) = \lim_{x\to 0^+}\frac{x - \sin x}{x\sqrt x\sin x} = \lim_{x \to 0^+}\frac{\frac16x^3 + o(x^3)}{x^{5/2} + o(x^{5/2})} = \lim_{x \to 0^+}\frac{\sqrt x}6 = 0$$ We used the fact that $$\sin x = x - \frac{x^3}{3!} + o(x^3)$$

2

Write it as $$\frac{x-\sin x}{x^{3/2}\sin x} \sim_{x\sim 0} \frac{x^3/3!}{x^{3/2}\,x} = \frac{1}{3!}x^{1/2}\longrightarrow_{x\to 0}0.$$

1

If you use L'hopital's rule two more times, you will reach the answer. Though using the finite expansion sin(x) ~ x - x^3/3! makes it a lot more easier. The limit will be zero. (Edited).

4

$$\lim_{n\to\infty}n\bigg(1-\frac{1}{\ln(n)}\bigg)^n=\lim_{n\to\infty}n\bigg[\bigg(1-\frac{1}{\ln(n)}\bigg)^{\ln(n)}\bigg]^{\frac{n}{\ln(n)}}=\lim_{n\to\infty}\frac{n}{e^{\frac{n}{\ln(n)}}}=0$$ Where I used the fact that $$\lim_{a_n\to\infty}(1-\frac{1}{a_n})^{a_n}=e^{-1}$$ and for $\ n\to\infty$ $$\ n>>\ln(n)$$

4

We have $$n\left(1-\frac{1}{\ln(n)}\right)^n=n\exp\left(n\ln\left(1-\frac{1}{\ln(n)}\right)\right)\sim_\infty n\exp\left(-\frac{n}{\ln n}\right)\xrightarrow{n\to\infty}0$$ and the last limit can be proved using the L'Hôpital's rule.

0

In general the relationship between position $x(t)$ and velocity $v(t)$ is given by $v(t) = \frac{dx(t)}{dt}$. If given the velocity $v(t)$ initially then one can find the position by $x(t) = \int v(t)dt + c$ where $c$ is a constant, determined by initial conditions [Hint: your initial condition in this case is that $x(0.002) = 2$]. Now, do you understand ...

1

Another great book is Introduction to Calculus and Classical Analysis by Omar Hijab. It contains many solved problems to show you how the reasoning is like and is well written. http://www.amazon.com/Introduction-Calculus-Classical-Undergraduate-Mathematics/dp/1441994874

0

Set $x=\dfrac1y$ $$I=\int_0^\infty\frac{\ln x}{1+x^2}dx=-\int_\infty^0\frac{\ln(y^{-1})}{1+y^2}dy$$ $$=\int_\infty^0\frac{\ln(y)}{1+y^2}dy$$ $$=-\int_0^\infty\frac{\ln(y)}{1+y^2}dy=-I$$

1

Use the Liebniz integral rule to get the result. That is, $$\frac{d}{dt}\int_{a(t)}^{b(t)}F(w,t)dw = \int_{a(t)}^{b(t)}\frac{\partial F}{\partial t}(w,t)dw + F(b(t),t)\frac{d b}{dt}(t)-F(a(t),t)\frac{d a}{dt}$$ In this case you have $$b(t)=\log^3(x(t))=4t^2 \cos(2+6t)$$ $$a(t) = \exp(4y(t))=\exp(4\log(2r+7\exp(5(t))))$$ and F(w,t) = ... 0 HINT: Set x=\tan y \implies x=0,y=\arctan 0=0 x=\infty\implies y=\dfrac\pi2 Finally use \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx 0 The given partitions (a_1, \dots, a_k) are all given by \begin{align*} a_i &= \begin{cases} a_1 & \text{if i \leq p}; \\ a_1 + 1 & \text{if i > p} \end{cases} \end{align*} for p = 1, \dots, k. Count the number of pairs (a_1, p) for which the partition sums to n. 1 HINT: As an alternative to an inductive approach: for each possible choice of k (the number of terms), there is exactly one choice of a_1 that allows the inequalities to be satisfied and the sum of the a_i’s to be n. 2 For a given positive integer n, let p_i(n) = (a_1, a_2, \ldots, a_k) be a given partition of n that satisfies the criteria. For each such partition p_i(n), how many ways are there to generate a unique partition p_i(n+1)? Is there a bijection? 0 I noticed that you had for n = 2 the solutions as 1+1, 2+0 but 2+0 cannot be a solution based on the problem constraints. Edit: My bad, as I was working through 2+0 did not make sense but 2 is clearly a solution. But I was able to prove this by thinking for each n, how many ways can you sum to n with k numbers (knowing the largest possible k is k =n where ... 1 I am following your idea.A = \{a \le x < x_0 | f(x) = 0 \}B = \{x_0 < x \le b | f(x) = 0 \}$$Let c = \sup A and let d = \inf B (\sup B doesn't work here, since it's always b). Note c<x_0<d Step 1: prove f(x) > 0 for x \in (c, d) we will use contradiction. Assume f(x) < 0, for some c<x<d, by IVT ... 3 Derivatives have the intermediate value property: if F is differentiable on [a, b], then F', on [a, b] takes on every value between F'(a) and F'(b). Suppose that a function f with the derivative you wrote really did exist. Look at a = -\frac{1}{\ln 3}; there, you have$$ f'(a) = \frac{e^{-\ln 3} + 1}{e^{-\ln 3}} = \frac{4/3}{1/3} = 4. $$... 1 Let \phi(x)=1+e^{-1/x}, x\ne0, \phi(0)=0. If x>0 then 1<\phi(x)<2. On the other hand \lim_{x\to0^-}\phi(x)=+\infty. Let a>0 be such that \phi(x)>2 if -a\le x<0. Then \phi does not have the mean value property on [a,\infty) and therefore it is not the derivative of any differentiable function. 4 You can integrate the function to get$$f(x)=\begin {cases}x+\frac 1{2x^2}+c_1&x \lt 0 \\x+\frac 1{2x^2}+c_2&x \gt 0 \end {cases}$$Note that the constants of integration may be different on the two sections of the domain. You have a minimum (not maximum) at x=1 The function blows up to +\infty approaching 0 from either side, which may be ... 0 We only need to show f is differentiable at each x\in I. For interior point x\in I, then \exists [a,b]\subset I such that x\in[a,b], then we apply the assumption. If x is end point of the interval, say the left endpoint, then just consider [x,b]\subset I. 2 Though tagged as "Calculus", I want add an algebraic solution$$(n+2)(n+1)\binom nr\frac1{(r+2)}=(r+1)\binom{n+2}{r+2}=[(r+2)-1]\binom{n+2}{r+2}=(n+2)\binom{n+1}{r+1}-\binom{n+2}{r+2}\implies(n+2)(n+1)\sum_{r=1}^n\binom nr\frac1{r+2}=(n+2)\sum_{r=1}^n\binom{n+1}{r+1}-\sum_{r=1}^n\binom{n+2}{r+2}$$... 2$$ \begin{align} \int_{0.5\pi}^{\pi}\frac{1}{1-\cos x}dx &=& \int_{0.5\pi}^{\pi}\frac{1}{2\sin^2\left(\frac{x}{2}\right)}dx\\ \end{align} $$Here I used$$ \cos(2x) = 1-2\sin^2x\implies \cos(x) = 1 -2\sin^2\left(\frac{x}{2}\right) $$using the sub x= 2u$$ \int \csc^2u du = -\cot u + C $$thus$$ \begin{align} -\cot u + C &=& ...

1

$$\frac{d}{dx}{\int_0^x{f(t})\,dt}=f(x)$$ This is a simplification of the general form of the Leibniz integral rule: $$\frac{d}{dx}{\int_{a(x)}^{b(x)}{f(x,t})\,dt}=f(x,b(x)).b'(x)-f(a,x(a)).a'(x)+\int_{a(x)}^{b(x)}{f_x(x,t})\,dt$$ The simplifications occurs, because the derivative of the upper limit equals to 1, the negative part equals to $0$ (since ...

1

Notice you're asked to find $f^{(n)}(1)$, not $f^{(n)}(x)$. A series expansion around $a=1$ is a nice way to do this without taking any derivatives. If you can find a power series representation of the form $f(x) = \sum_{n=0}^\infty c_n (x-a)^n$ with positive radius of convergence, then $c_n = \frac{f^{(n)}(a)}{n!}$. Therefore $f^{(n)}(a) = c_n n!$. ...

4

Multiply numerator and denominator by $1+\cos x$ to get \begin{align}\int_{0.5\pi}^\pi \frac{(1+\cos x)\,dx}{1-\cos^2 x}& = \int_{0.5\pi}^\pi \frac{(1+\cos x)\,dx}{\sin^2 x} \\ \\ &= \int \csc^2 x \,dx + \int \frac{\cos x\,dx}{\sin^2 x}\\ \\ & = -\cot x + \int \cot x\csc x\,dx\\ \\ &=-\cot x -\csc x+C\end{align}

1

Your equality is the direct consequence of the fact that $$\left(\int_0^y f(t)dt\right)'= f(y),$$ a well known fact that holds for all $y\in [0,b)$ if $f$ is continuous on some $[0,b)$ (in fact, this is true even a slighlty larger class of functions).

1

Since for any $a\neq 0$: $$\frac{1}{x(a-x)} = \frac{1}{a}\left(\frac{1}{x}+\frac{1}{a-x}\right)\tag{1}$$ we have: $$\frac{d^n}{dx^n}\frac{1}{x(a-x)} = \frac{n!}{a}\left(\frac{(-1)^n}{x^{n+1}}+\frac{1}{(a-x)^{n+1}}\right).\tag{2}$$

2

hint: $$\dfrac{2}{x(2-x)} = \dfrac{1}{x} - \dfrac{1}{x-2}$$ taking derivatives of $x^{-1}$ and $(x-1)^{-1}$ should be much easier than the quotient rule. $\bf {edit:}$ you may not have done the taylor series yet but if you have, we can compute $f^{n}(1)$ without taking any derivatives at all. here is how you do it. we make change of variable \$x = 1 + ...

1

Since: $$\int_{0}^{1}\frac{dt}{1-xt}=-\frac{\log(1-x)}{x}$$ and: $$I=\int_{0}^{1}\frac{\log x}{x-1}\,dx = -\int_{0}^{1}\frac{\log(1-x)}{x}\,dx$$ we have: $$I = \iint_{(0,1)^2}\frac{1}{1-xt}\,dt\,dx = \sum_{n\geq 0}\iint_{(0,1)^2}(xt)^n\,dt\,dx = \sum_{n\geq 0}\frac{1}{(n+1)^2}=\color{red}{\zeta(2)}$$ as wanted.

Top 50 recent answers are included