Could someone help me complete this or check if my reasoning so far is correct? I'm stuck at finding the oblique asymptote:
Write the Taylor polynomial and the Peano form of the remainder of $f(x) = (x-1)arctan x$.
$n = 2$
$x_0 = 0$
Also find the asymptotes.
$f'(x) = \frac{d}{dx}(x-1)\arctan x+ \frac{d}{dx}\arctan x(x-1)$ = $\arctan x + \frac{x-1}{1+x^2}$
$f''(x) = \frac{d}{dx}(\arctan x + \frac{x-1}{1+x^2})$ = $\frac{1}{1+x^2}+\frac{1*(1+x^2)-(2x)(x-1)}{(1+x^2)^2}$ = $\frac{1}{1+x^2}-\frac{x^2-2x-1}{(1+x^2)^2}$
Taylor Polynomial: $(fx_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2}(x-x_0)^2+o((x-x_0)^2$
(Here, I am not sure I understood how the Peano remainder works. Do I write it once with n = 2 or do I do it twice for n = 1, n = 2?)
Where:
$f(x_0) = -1*0 = 0$
$f'(x_0) = 0-1 = -1$
$f''(x_0) = 1+1 = 2$
So, $f(x) = x^2-x+o(x)^2$
Asymptotes: Domain of function is R. No vertical asymptote.
No horizontal asymptote: limit for ($x-> \infty$) is infinity.
I look for an oblique asymptote:
m is the limit of $\frac{f(x)}{x}$, which gives an indeterminate form. Applying Hopital rule:
$\frac{\arctan x+\frac{x-1}{1+x^2}}{1} = \frac{pi}{2} + \frac{\infty}{\infty}$
I solve the indeterminate form, rewriting the fraction: $\frac{x(1-\frac{1}{x})}{x^2+(\frac{1}{x^2}+1)}$ = $\frac{1-\frac{1}{x}}{x(\frac{1}{x^2}+1)}$ = $\frac{1-0}{\infty}$ = 0
So, $m = \frac{pi}{2}$
q is the infinite limit of $f(x) - mx$
$q = (x-1)\arctan x- \frac{pi}{2}x$ which is ($\infty - \infty$)
I am stuck here, as I have no idea how to solve the form: if anyone could help me, that'd be greatly appreciated.