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4

May be helpful to change variables to let $u = x-1 \to 0$ then the limit becomes $$ \begin{split} \lim_{x \to 1} \frac{\ln x}{x^2-1} &= \lim_{x \to 1} \frac{\ln x}{(x+1)(x-1)} \\ &= \lim_{u \to 0} \frac{\ln (1+u)}{u(u+2)} \\ &= \lim_{u \to 0} \frac{u - \frac{u^2}{2}+\frac{u^3}{3} \ldots}{u(u+2)} \\ &= \lim_{u \to 0} \frac{1 - \frac{u}{2}+\...


2

The constant term is $f(0)=\ln 2>0.$ $\textbf{Update:}$ The series for $f(x)=\ln \, (2+x)$ is $$f(x)=\underbrace{\ln 2}_\textrm{const. term} + \frac{1}{2} x -\frac{1}{4}x^2 + \cdots.$$ The constant term is the first term and it is positive.


2

There's a sign error in your $x^3$ coefficient. With $\equiv$ denoting equality up to $x^5$ terms, $$\exp(x\cos x)\equiv\exp\left(x-\frac{x^3}{2}+\frac{x^5}{24}\right)\\\equiv\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\right)\left(1-\frac{x^3}{2}\right)\left(1+\frac{x^5}{24}\right)\\\equiv\left(1+x+\frac{x^2}{2}-\frac{x^3}{3}-\frac{...


1

Since $\ln(x+1)=x+O(x^2)$, $\ln(x)=\ln\bigl((x-1)+1\bigr)=x-1+O\bigl((x-1)^2\bigr)$. And since $\sin(x)=x+O(x^2)$,\begin{align}\sin(\pi x)&=-\sin(\pi x-\pi)\\&=-\sin\bigl(\pi(x-1)\bigr)\\&=-\pi(x-1)+O\bigl((x-1)^2\bigr).\end{align}So\begin{align}\frac{\ln(x)-\sin(\pi x)}{\sqrt{x-1}}&=\sqrt{x-1}\frac{\ln(x)-\sin(\pi x)}{x-1}\\&\to0\times(1-...


1

Hint: Look up multivariable Taylor series. You will need the matrices of first and second partial derivatives (aka the Jacobian and the Hessian). Let me help. $F_x(x,y)=2e^y(x-1)+1, F_y(x,y)=e^y(x-1)^2, F_{xx}(x,y)=2e^y, F_{yy}(x,y)=e^y(x-1)^2$ and $F_{xy}(x,y)=2xe^y-2e^y$. Now plug in the two points: $F_x(0,0)=-1, F_y(0,0)=1, F_{xx}(0,0)=2, F_{yy}(0,0)...


1

You need expansion around the point you want to find limit in. $\ln x = \ln(1 + (x - 1)) = (x - 1) + o(x - 1)$, $x^2 - 1 = (x - 1) (x + 1) = (x - 1) \cdot 2 + o(x - 1)$ (all $o$ with base $x \to 1$). So the expression under limit is $\frac{(x - 1) + o(x - 1)}{2(x - 1) + o(x - 1)} = \frac{1}{2} + o(1)$.


1

Use the standard series $$\tan ^{-1}(t)=t-\frac{t^3}{3}+\frac{t^5}{5}+O\left(t^{7}\right)$$ $$\frac{t}{\tan ^{-1}(t)}=\frac {t} {t-\frac{t^3}{3}+\frac{t^5}{5}+O\left(t^{7}\right) }=\frac {1} {1-\frac{t^2}{3}+\frac{t^4}{5}+O\left(t^{6}\right) }$$ Use long division $$\frac{t}{\tan ^{-1}(t)}=1+\frac{t^2}{3}-\frac{4 t^4}{45}+O\left(t^6\right)=1+\frac{t^2}{3}-\...


1

Let $f_N(z) = \sum_{n=0}^N c_n z^n$ with $c_n $ real, if its $N$ zeros $a_{N,1},\ldots,a_{N,N}$ are real and simple there is $r_{N+1} \le 1$ such that for any $c_{N+1}\in [- r_{N+1},r_{N+1}]$, $f_{N+1}(z) = c_{N+1} z^{N+1}+f_N(z)$ has sign changes and $f_{N+1}'(z)$ stays of constant sign near $a_{N,j}$, so that $a_{N+1,j} \approx a_{N,j}$ is real and simple....


1

All values and derivatives are Taylor expanded at $t$ so they can be compared. You could just as well start with a polynomial or power series, $y(t+s)=\sum a_ks^k$, $y'(t+s)=\sum ka_ks^{k-1}$ and then start comparing coefficients starting at the lowest degree. Note that the resulting method is extremely unstable, as local errors, truncation or floating ...


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