User:X42bn6/Sandbox

$$ \begin{align} y &= \displaystyle\frac{\sqrt{5+x^2}}{3-x} \\ \displaystyle\frac{d}{dx}\left(\sqrt{5+x^2}\right) &= \displaystyle\frac{2x}{2\sqrt{5+x^2}}\\ &= \displaystyle\frac{x}{\sqrt{5+x^2}} \\ \displaystyle\frac{d}{dx}(3-x) &= -1 \\ \displaystyle\lim_{x\to\pm\infty}\displaystyle\frac{\sqrt{5+x^2}}{3-x} &= \displaystyle\lim_{x\to\pm\infty}-\displaystyle\frac{x}{\sqrt{5+x^2}} && \mbox{(1), iif limit exists} \\ &= \displaystyle\lim_{x\to\pm\infty}\mp\sqrt{\displaystyle\frac{x^2}{5+x^2}} && \mbox{Remember negative values; alternatively, consider the positive and negative cases separately} \\ &= \displaystyle\lim_{x\to\pm\infty}\mp\sqrt{\displaystyle\frac{5+x^2-5}{5+x^2}} \\ &= \displaystyle\lim_{x\to\pm\infty}\mp\sqrt{1-\displaystyle\frac{5}{5+x^2}} \\ &\to \mp 1 && \mbox{Also satisfies (1)} \\ y &= \displaystyle\frac{\sqrt{5+x^2}}{3-x} \\ & \vdots \\ x &= \displaystyle\frac{3y^2\pm\sqrt{14y^2-5}}{y^2-1} \\ &= \displaystyle\frac{3(y^2-1)+3\pm\sqrt{14y^2-5}}{y^2-1} \\ &= 3+\displaystyle\frac{\pm\sqrt{14y^2-5}}{y^2-1} \\ &= 3\pm\sqrt{\displaystyle\frac{14y^2-5}{\left(y^2-1\right)^2}} \\ &= 3\pm\sqrt{\displaystyle\frac{14z-5}{\left(z-1\right)^2}} && z=y^2 \\ \exists\mbox{constants } A, B\,.\,\\ \displaystyle\frac{14z-5}{\left(z-1\right)^2} &= \displaystyle\frac{A}{\left(z-1\right)^2}+\displaystyle\frac{B}{z-1} && \mbox{Partial fractions} \\ \displaystyle\lim_{z\to\pm\infty}\displaystyle\frac{14z-5}{\left(z-1\right)^2} &= \displaystyle\lim_{z\to\pm\infty}\displaystyle\frac{A}{\left(z-1\right)^2}+\displaystyle\lim_{z\to\pm\infty}\displaystyle\frac{B}{z-1} && \mbox{Limit of sums is sum of limits}\\ &= 0 \\ \therefore\displaystyle\lim_{y\to\pm\infty} x &= \displaystyle\lim_{y\to\pm\infty}3\pm\sqrt{0} && \mbox{Note that }z\geq 0\\ &= 3 \end{align} $$