## Grenseverdiløysingar for nokre kvadratrotuttrykk

Juni 2016

$$f(x) = \sqrt{x^2+x+1} - x$$

$$f(x) = \sqrt{x^2+x+1} - x$$

$$x\to{-\infty}$$

$$x^2+x+1 \to\infty$$

$$\sqrt{x^2+x+1}\to\infty$$

$$-x\to\infty$$

$$f(x)\to\infty$$

$$f(x) = \sqrt{x^2+x+1} - x$$

$$x\to{\infty}$$

$$f(x) = \big[\sqrt{x^2+x+1} - x\big] \frac{\sqrt{x^2+x+1} + x}{\sqrt{x^2+x+1} + x}$$

$$f(x) = \frac{\sqrt{x^2+x+1}^2 - x^2}{\sqrt{x^2+x+1} + x}$$

$$f(x) = \frac{x^2+x+1 - x^2}{\sqrt{x^2+x+1} + x^2}$$

$$f(x) = \frac{x+1}{\sqrt{x^2+x+1} + x}$$

$$f(x) = \frac{\frac xx+\frac1x}{\frac{\sqrt{x^2+x+1}}x + \frac xx}$$

$$f(x) = \frac{1+\frac1x}{\sqrt{\frac{x^2}{x^2}+\frac x{x^2}+\frac1{x^2}} + 1}$$

$$f(x) = \frac{1+\frac1x}{\sqrt{1+\frac1x+\frac1{x^2}} + 1}$$

$$f(x) \to \frac12$$

$$f(x) = \sqrt{x^2+x+1} - x$$
\begin{align*} \lim_{x\to\infty} f(x) = \frac12 \\ \lim_{x\to{-\infty}} f(x) = \infty \end{align*}

$$g(x) = \frac1{\sqrt{x^2-x} + x}$$

$$g(x) = \frac1{\sqrt{x^2-x} + x}$$

$$x\to{\infty}$$

$$g(x) = \frac{\sqrt{x^2-x} - x} {\big[\sqrt{x^2-x} + x\big]\cdot\big[\sqrt{x^2-x} - x\big]}$$

$$g(x) = \frac{\sqrt{x^2-x} - x}{\sqrt{x^2-x}^2 - x^2}$$

$$g(x) = \frac{\sqrt{x^2-x} - x}{x^2-x - x^2}$$

$$g(x) = \frac{\sqrt{x^2-x} - x}{-x}$$

$$g(x) = \frac{\frac{\sqrt{x^2-x}}x - \frac xx}{-\frac xx}$$

$$g(x) = \frac{\sqrt{\frac{x^2}{x^2}-\frac{x}{x^2}} - \frac xx}{-\frac xx}$$

$$g(x) = \frac{\sqrt{1-\frac{1}{x}} - 1}{-1}$$

$$g(x) \to 0$$

$$g(x) = \frac1{\sqrt{x^2-x} + x}$$

$$x\to{-\infty}$$

$$g(x) = \frac{\sqrt{x^2-x} - x} {\big[\sqrt{x^2-x} + x\big]\cdot\big[\sqrt{x^2-x} - x\big]}$$

$$g(x) = \frac{\sqrt{x^2-x} - x}{\sqrt{x^2-x}^2 - x^2}$$

$$g(x) = \frac{\sqrt{x^2-x} - x}{x^2-x - x^2}$$

$$g(x) = \frac{\sqrt{x^2-x} - x}{-x}$$

$$g(x) = \frac{\frac{\sqrt{x^2-x}}x - \frac xx}{-\frac xx}$$

$$g(x) = \frac{-\sqrt{\frac{x^2}{x^2}-\frac{x}{x^2}} - \frac xx}{-\frac xx}$$

$$g(x) = \frac{-\sqrt{1-\frac{1}{x}} - 1}{-1}$$

$$g(x) \to 2$$

$$g(x) = \frac1{\sqrt{x^2-x} + x}$$

\begin{align*} \lim_{x\to\infty} g(x) = 0 \\ \lim_{x\to{-\infty}} g(x) = 2 \end{align*}

$$\lim_{x\to\infty} \sqrt{3x^2 + 1} - 2x$$

$$\lim_{x\to{-\infty}} \sqrt{3x^2 + 1} - 2x$$