Answer:
32
Step-by-step explanation:
We can apple L'Hopital's Rule to solve this problem:
[tex]\lim _{x\to \:\:0}\left(\frac{e^{8x}-8x-1}{x^2}\right)\\=> \lim _{x\to \:0}\left(\frac{8e^{8x}-8}{2x}\right)\\\\=> \lim _{x\to \:0}\left(\frac{4\left(e^{8x}-1\right)}{x}\right)\\\\=> \lim _{x\to \:0}\left(\frac{32e^{8x}}{1}\right)\\\\=>\frac{32e^{8\cdot \:0}}{1}\\\\=> 32[/tex]
Hope this helps!
