No LIMITs to the amount of limit problems!

Since plugging in $0$ here gives us a $\cfrac{0}{0}$ indeterminate form, we can use L'Hopital's Rule and differentiate the top and the bottom of the fraction in the limit: $\begin{aligned} \lim\limits_{x\to 0} \frac{17^x - 1}{x} \implies \lim\limits_{x\to 0} \frac{\frac{d}{dx}(17^x - 1)}{\frac{d}{dx}(x)} &= \lim\limits_{x\to 0} \frac{17^x \ln(17) }{1} \\ &= \lim\limits_{x\to 0} 17^x \ln(17) \\ &= 17^0 \ln(17) = \green{\boxed{\ln(17)}} = \green{\boxed{2.83321334406\dots}}\end{aligned}$

This is your classical L'Hôpital problem; differentiating both top and bottom will result in:

$\displaystyle \frac{d}{dx}17^x = \ln(17) 17^x$

$\displaystyle \frac{d}{dx}x = 1$

Plug in zero to the numerator, we get:

$\boxed{\ln(17)}$