Answer:
40.5
Step-by-step explanation:
Given the limit of the function expressed as:
[tex]\lim_{x \to 0} \frac{e^{9x}-1-9x}{x^2} \\[/tex]
Step 1: Substitute x = 0 into the function:
[tex]\lim_{x \to 0} \frac{e^{9x}-1-9x}{x^2} \\= \frac{e^{9(0)}-1-9(0)}{(0)^2}\\= \frac{{1-1-0}}{0}\\= \frac{0}{0} (ind)[/tex]
Step 2: Apply L'hopital rule
[tex]\lim_{x \to 0} \frac{d/dx(e^{9x}-1-9x)}{d/dx(x^2)} \\= \lim_{x \to 0} \frac{(9e^{9x}-9)}{(2x)}\\= \frac{(9e^{9(0)}-9)}{(2(0))}\\= \frac{9-9}{0} \\= \frac{0}{0} (ind)[/tex]
Step 3: Apply L'hopital rule again
[tex]\lim_{x \to 0} \frac{d/dx(9e^{9x}-9)}{d/dx(2x)}\\ = \lim_{x \to 0} \frac{81e^{9x}}{2} \\=\frac{81e^{0}}{(2)}\\= \frac{81}{2} \\= 40.5[/tex]
Hence the limit of the function is 40.5
