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Find the indicated limit. Note that L'Hopital's rule may not apply or may require multiple applications.

lim e^9x -1-9x/x^2
x→0

Sagot :

abidemiokin

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

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