$$\lim_{x\to\infty} \sqrt{3x^2 + 1} - 2x$$
$$\lim_{x\to{-\infty}} \sqrt{3x^2 + 1} - 2x$$
Hans Georg Schaathun
Juni 2016
$$f(x) = \sqrt{x^2+x+1} - x$$
$$x\to{-\infty}$$
$$x^2+x+1 \to\infty$$
$$\sqrt{x^2+x+1}\to\infty$$
$$-x\to\infty$$
$$f(x)\to\infty$$
$$f(x) = \sqrt{x^2+x+1} - x$$
$$x\to{\infty}$$
$$f(x) = \big[\sqrt{x^2+x+1} - x\big] \frac{\sqrt{x^2+x+1} + x}{\sqrt{x^2+x+1} + x} $$
$$f(x) = \frac{\sqrt{x^2+x+1}^2 - x^2}{\sqrt{x^2+x+1} + x}$$
$$f(x) = \frac{x^2+x+1 - x^2}{\sqrt{x^2+x+1} + x^2}$$
$$f(x) = \frac{x+1}{\sqrt{x^2+x+1} + x}$$
$$f(x) = \frac{\frac xx+\frac1x}{\frac{\sqrt{x^2+x+1}}x + \frac xx}$$
$$f(x) = \frac{1+\frac1x}{\sqrt{\frac{x^2}{x^2}+\frac x{x^2}+\frac1{x^2}} + 1}$$
$$f(x) = \frac{1+\frac1x}{\sqrt{1+\frac1x+\frac1{x^2}} + 1}$$
$$f(x) \to \frac12$$
$$g(x) = \frac1{\sqrt{x^2-x} + x}$$
$$g(x) = \frac1{\sqrt{x^2-x} + x}$$
$$x\to{\infty}$$
$$g(x) = \frac{\sqrt{x^2-x} - x} {\big[\sqrt{x^2-x} + x\big]\cdot\big[\sqrt{x^2-x} - x\big]} $$
$$g(x) = \frac{\sqrt{x^2-x} - x}{\sqrt{x^2-x}^2 - x^2}$$
$$g(x) = \frac{\sqrt{x^2-x} - x}{x^2-x - x^2}$$
$$g(x) = \frac{\sqrt{x^2-x} - x}{-x}$$
$$g(x) = \frac{\frac{\sqrt{x^2-x}}x - \frac xx}{-\frac xx}$$
$$g(x) = \frac{\sqrt{\frac{x^2}{x^2}-\frac{x}{x^2}} - \frac xx}{-\frac xx}$$
$$g(x) = \frac{\sqrt{1-\frac{1}{x}} - 1}{-1}$$
$$g(x) \to 0$$
$$g(x) = \frac1{\sqrt{x^2-x} + x}$$
$$x\to{-\infty}$$
$$g(x) = \frac{\sqrt{x^2-x} - x} {\big[\sqrt{x^2-x} + x\big]\cdot\big[\sqrt{x^2-x} - x\big]} $$
$$g(x) = \frac{\sqrt{x^2-x} - x}{\sqrt{x^2-x}^2 - x^2}$$
$$g(x) = \frac{\sqrt{x^2-x} - x}{x^2-x - x^2}$$
$$g(x) = \frac{\sqrt{x^2-x} - x}{-x}$$
$$g(x) = \frac{\frac{\sqrt{x^2-x}}x - \frac xx}{-\frac xx}$$
$$g(x) = \frac{-\sqrt{\frac{x^2}{x^2}-\frac{x}{x^2}} - \frac xx}{-\frac xx}$$
$$g(x) = \frac{-\sqrt{1-\frac{1}{x}} - 1}{-1}$$
$$g(x) \to 2$$
$$g(x) = \frac1{\sqrt{x^2-x} + x}$$
$$\lim_{x\to\infty} \sqrt{3x^2 + 1} - 2x$$
$$\lim_{x\to{-\infty}} \sqrt{3x^2 + 1} - 2x$$