Substituting x = 0 directly gives the indeterminate form 0/0, so you can use l'Hopital's rule (and you'll have to do that twice):
[tex]\displaystyle\lim_{x\to0}\frac{e^{6x}-6x-1}{x^2}=\lim_{x\to0}\frac{6e^{6x}-6}{2x}=\lim_{x\to0}\frac{36e^{6x}}2[/tex]
The limand is continuous everywhere, so you can plug in x = 0 to get 36•1/2 = 18.
