Answer :

The limit of the expression [tex]\(\frac{1}{h} \int_{0}^{2h} f(x) \, dx\) as \(h\) approaches 0 is equal to \(2f(0)\).[/tex]

To evaluate the limit [tex]\(\lim_{{h \to 0}} \frac{1}{h} \int_{0}^{2h} f(x) \, dx\)[/tex], we can use the Mean Value Theorem for integrals.

The Mean Value Theorem for integrals states that if a function [tex]\(f(x)\)[/tex] is continuous on the closed interval [tex]\([a, b]\)[/tex]and differentiable on the open interval [tex]\((a, b)\),[/tex] then there exists a value c in the open interval (a, b) such that:

[tex]\[\int_{a}^{b} f(x) \, dx = (b - a) \cdot f(c)\][/tex]

In our case, the function [tex]\(f(x)\)[/tex]is differentiable on [tex]\((-1, 1)\)[/tex], which includes the interval [tex]\((0, 2h)\)[/tex] for small values of[tex]\(h\)[/tex]. So, we can apply the Mean Value Theorem for integrals to the integral [tex]\(\int_{0}^{2h} f(x) \, dx\)[/tex]on the interval [tex]\([0, 2h]\).[/tex]

Let's set [tex]\(a = 0\) and \(b = 2h\):[/tex]

[tex]\[\int_{0}^{2h} f(x) \, dx = (2h - 0) \cdot f(c)\][/tex]

where c is some value in the open interval[tex]\((0, 2h)\).[/tex]

Now, divide both sides by [tex]\(2h\)[/tex] to find the average value of [tex]\(f(x)\) on the interval \((0, 2h)\):[/tex]

[tex]\[\frac{1}{2h} \int_{0}^{2h} f(x) \, dx = f(c)\][/tex]

Finally, take the limit as h approaches 0:

[tex]\[\lim_{{h \to 0}} \frac{1}{h} \int_{0}^{2h} f(x) \, dx = \lim_{{h \to 0}} 2f(c) = 2f(0)\][/tex]

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