Question:

Step 1: Express $f(x)$ as a derivative of a sum. The given function is $f(x) = \displaystyle\lim_{n \to \infty} \sum_{r=1}^{n} \dfrac{1}{2^{2r}} \sec^2 \dfrac{x}{2^r}$. Observe that the term $\dfrac{1}{2^{2r}} \sec^2 \dfrac{x}{2^r}$ is the derivative of $\dfrac{\tan(x/2^r)}{2^r}$ with respect to $x$: $$ \dfrac{d}{dx}\left(\dfrac{\tan(x/2^r)}{2^r}\right) = \dfrac{1}{2^r} \cdot \sec^2\left(\dfrac{x}{2^r}\right) \cdot \dfrac{1}{2^r} = \dfrac{1}{2^{2r}}\sec^2\dfrac{x}{2^r} $$ Therefore, $f(x)$ can be written as: $$ f(x) = \dfrac{d}{dx}\left(\lim_{n \to \infty} \sum_{r=1}^{n} \dfrac{\tan(x/2^r)}{2^r}\right) $$ Step 2: Evaluate the sum using a telescoping identity. We use the trigonometric identity $\tan A = \cot A - 2\cot(2A)$. Let $A = x/2^r$. Then, $\tan(x/2^r) = \cot(x/2^r) - 2\cot(x/2^{r-1})$. Dividing by $2^r$, we get: $$ \dfrac{\tan(x/2^r)}{2^r} = \dfrac{\cot(x/2^r)}{2^r} - \dfrac{2\cot(x/2^{r-1})}{2^r} = \dfrac{\cot(x/2^r)}{2^r} - \dfrac{\cot(x/2^{r-1})}{2^{r-1}} $$ This is a telescoping sum: $$ \sum_{r=1}^{n} \dfrac{\tan(x/2^r)}{2^r} = \sum_{r=1}^{n} \left(\dfrac{\cot(x/2^r)}{2^r} - \dfrac{\cot(x/2^{r-1})}{2^{r-1}}\right) $$ $$ = \left(\dfrac{\cot(x/2)}{2} - \dfrac{\cot x}{1}\right) + \left(\dfrac{\cot(x/4)}{4} - \dfrac{\cot(x/2)}{2}\right) + \dots + \left(\dfrac{\cot(x/2^n)}{2^n} - \dfrac{\cot(x/2^{n-1})}{2^{n-1}}\right) $$ $$ = \dfrac{\cot(x/2^n)}{2^n} - \cot x $$ Step 3: Evaluate the limit and determine the closed form of $f(x)$. Now, we take the limit as $n \to \infty$: $$ \lim_{n \to \infty} \left(\dfrac{\cot(x/2^n)}{2^n} - \cot x\right) = \lim_{n \to \infty} \dfrac{\cos(x/2^n)}{2^n \sin(x/2^n)} - \cot x $$ As $n \to \infty$, $x/2^n \to 0$. Using the small angle approximation $\sin \theta \approx \theta$ and $\cos \theta \approx 1$ for small $\theta$: $$ \lim_{n \to \infty} \dfrac{1}{2^n (x/2^n)} - \cot x = \lim_{n \to \infty} \dfrac{1}{x} - \cot x = \dfrac{1}{x} - \cot x $$ Therefore, the function $f(x)$ is: $$ f(x) = \dfrac{d}{dx}\left(\dfrac{1}{x} - \cot x\right) = -\dfrac{1}{x^2} + \csc^2 x $$ Step 4: Compute $f\!\left(\dfrac{\pi}{2}\right)$. Substitute $x = \dfrac{\pi}{2}$ into the expression for $f(x)$: $$ f\!\left(\dfrac{\pi}{2}\right) = -\dfrac{1}{(\pi/2)^2} + \csc^2\left(\dfrac{\pi}{2}\right) = -\dfrac{4}{\pi^2} + 1 = 1 - \dfrac{4}{\pi^2} $$ Step 5: Compute $f'(x)$ and evaluate at $x = \dfrac{\pi}{2}$. First, find the derivative of $f(x)$: $$ f'(x) = \dfrac{d}{dx}\left(-x^{-2} + \csc^2 x\right) = 2x^{-3} + 2\csc x (-\csc x \cot x) = \dfrac{2}{x^3} - 2\csc^2 x \cot x $$ Now, substitute $x = \dfrac{\pi}{2}$: $$ f'\!\left(\dfrac{\pi}{2}\right) = \dfrac{2}{(\pi/2)^3} - 2\csc^2\left(\dfrac{\pi}{2}\right)\cot\left(\dfrac{\pi}{2}\right) = \dfrac{2 \cdot 8}{\pi^3} - 2(1)^2(0) = \dfrac{16}{\pi^3} $$ Step 6: Compute $\displaystyle\lim_{x \to 0^+} f(x)$. $$ \lim_{x\to 0^+} f(x) = \lim_{x\to 0^+}\left(\csc^2 x - \dfrac{1}{x^2}\right) = \lim_{x\to 0^+}\left(\dfrac{1}{\sin^2 x} - \dfrac{1}{x^2}\right) = \lim_{x\to 0^+}\dfrac{x^2 - \sin^2 x}{x^2 \sin^2 x} $$ Using the Taylor series expansion for $\sin x = x - \dfrac{x^3}{3!} + O(x^5)$: $$ \sin^2 x = \left(x - \dfrac{x^3}{6} + O(x^5)\right)^2 = x^2 - 2x\left(\dfrac{x^3}{6}\right) + O(x^6) = x^2 - \dfrac{x^4}{3} + O(x^6) $$ Substitute this into the limit expression: $$ \lim_{x\to 0^+} \dfrac{x^2 - (x^2 - \dfrac{x^4}{3} + O(x^6))}{x^2 (x^2 - \dfrac{x^4}{3} + O(x^6))} = \lim_{x\to 0^+} \dfrac{\dfrac{x^4}{3} + O(x^6)}{x^4 - \dfrac{x^6}{3} + O(x^8)} $$ Divide the numerator and denominator by $x^4$: $$ \lim_{x\to 0^+} \dfrac{\dfrac{1}{3} + O(x^2)}{1 - \dfrac{x^2}{3} + O(x^4)} = \dfrac{1/3}{1} = \dfrac{1}{3} $$ Step 7: Analyze if $f(x) = 0$ has any real roots in $(0, \pi)$. The function is $f(x) = \csc^2 x - \dfrac{1}{x^2} = \dfrac{x^2 - \sin^2 x}{x^2 \sin^2 x}$. For $x \in (0, \pi)$, $\sin x > 0$, so $x^2 \sin^2 x > 0$. Consider the numerator $x^2 - \sin^2 x = (x - \sin x)(x + \sin x)$. For $x > 0$, it is a fundamental inequality that $x > \sin x$. Therefore, for $x \in (0, \pi)$, $x - \sin x > 0$. Also, for $x \in (0, \pi)$, $x + \sin x > 0$. Thus, $x^2 - \sin^2 x > 0$ for all $x \in (0, \pi)$. Since both the numerator and denominator are positive for $x \in (0, \pi)$, it follows that $f(x) > 0$ for all $x \in (0, \pi)$. Consequently, $f(x) = 0$ has no real roots in the interval $(0, \pi)$.