232

$f\left(x\right)=\begin{cases} 3x-6{,}&kun\ x\le6\\ \frac{x^2-36}{x-6}&kun\ x>6 \end{cases}$
$\lim_{x\rightarrow6-}=3\cdot6-6=12$
$\lim_{x\rightarrow6+}=\frac{6^2-36}{6-6}=\frac{0}{0}$

pitää sieventää

$\frac{\left(x-6\right)\left(x+6\right)}{x-6}=x+6$
$\lim_{x\rightarrow6+}=6+6=12$

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