Answer:
Step-by-step explanation:
[tex]\lim_{x \to \ 0} \frac{e^{2x}-1 }{x} \\= \lim_{x \to \0} \frac{e^{2x}-1 }{2 x} Â *2\\\rightarrow2 log e\\\rightarrow 2[/tex]
The term arbitrarily close means that the value of f(x) is along the line of its point of intersection.
The equation that expresses the property that f(x) can be made arbitrarily close to 2 by taking x sufficiently close to 0, but not equal to 0 is [tex]\lim_{x \to 0} \frac{e^{2x} - 1}{x} = 2[/tex]
Given that:
[tex]f(x) =\frac{e^{2x} - 1}{x}[/tex]
To determine the equation that satisfies the given property, we simply take the limit of x to 0 and the equation must equal to 2.
So, we have:
[tex]\lim_{x \to 0} f(x) = 2[/tex]
Substitute [tex]f(x) =\frac{e^{2x} - 1}{x}[/tex]
[tex]\lim_{x \to 0} \frac{e^{2x} - 1}{x} = 2[/tex]
Hence, the equation that expresses the property that f(x) can be made arbitrarily close to 2 by taking x sufficiently close to 0, but not equal to 0 is [tex]\lim_{x \to 0} \frac{e^{2x} - 1}{x} = 2[/tex]
Read more about arbitrarily values at:
https://brainly.com/question/14662242
