Grenseverdisetningene

$\begin{array}{rcl}\underset{x\to 1}{\lim }3x & =& \underset{x\to 1}{\lim }3·\underset{x\to 1}{\lim }x\\ & =& 3·1\\ & =& 3\end{array}$

$\begin{array}{rcl}\underset{x\to 2}{\lim }\left(x^2+3x\right) & =& \underset{x\to 2}{\lim }{x}^2+\underset{x\to 2}{\lim }3x\\ & = & 2^2+3·2\\ & = & 4+6 \\ & =& 10\end{array}$

$\begin{array}{rcl}\underset{x\to 3}{\lim }\left(x^3+x^2-6^2\right) & =& \underset{x\to 3}{\lim }{x}^3+\underset{x\to 3}{\lim }{x}^2-\underset{x\to 3}{\lim }{6}^2\\ & =& 3^3+3^2-6^2 \\ & =& 27+9-36 \\ & =& 0\end{array}$

$\begin{array}{rcl}\underset{x\to 3}{\lim }\frac{5x-10}{x^2} & =& 5·\frac{\underset{x\to 3}{\lim }\left(x-2\right)}{\underset{x\to 3}{\lim }{x}^2}\\ & & \\ & =& 5·\frac{3-2}{3^2} \\ & & \\ & =& 5·\frac{1}{9}\\ & & \\ & =& \frac{5}{9} \end{array}$

$\begin{array}{rcl}\underset{x\to \infty }{\lim }\left(4x+3\right) & =& \underset{x\to \infty }{\lim }4x+\underset{x\to \infty }{\lim }3\end{array}$

Siden 4x går mot uendelig når x går mot uendelig, eksisterer ikke grenseverdien.

$\begin{array}{rcl}\underset{x\to \infty }{\lim }\frac{3}{x}& = & \underset{x\to \infty }{\lim }3·\frac{1}{x}\\ & & \\ & =& 3·\underset{x\to \infty }{\lim }\frac{1}{x}\\ & & \\ & =& 3·0\\ & & \\ & =& 0\end{array}$