Question:

Step 1: Since $f$ is a continuous function on $\mathbb{R}$, we can determine $f(0)$ by evaluating the limit of $f\left(\frac{1}{n^4}\right)$ as $n \to \infty$. As $n \to \infty$, $\frac{1}{n^4} \to 0$. Therefore, $$f(0) = \lim_{n\to\infty} f\left(\frac{1}{n^4}\right)$$ Step 2: The given expression for $f\left(\frac{1}{n^4}\right)$ is: $$f\left(\frac{1}{n^4}\right) = \frac{(\sin e^{-n^2})e^{-n^2} + \frac{n^2}{n^2+1}}{1}$$ This simplifies to: $$f\left(\frac{1}{n^4}\right) = (\sin e^{-n^2})e^{-n^2} + \frac{n^2}{n^2+1}$$ Step 3: Evaluate the limit of the first term as $n \to \infty$: As $n \to \infty$, $e^{-n^2} \to 0$. Let $x = e^{-n^2}$. Then as $n \to \infty$, $x \to 0$. The limit of the first term becomes: $$\lim_{n\to\infty} (\sin e^{-n^2})e^{-n^2} = \lim_{x\to 0} (\sin x)x$$ Since $\lim_{x\to 0} \sin x = 0$ and $\lim_{x\to 0} x = 0$, their product is: $$\lim_{x\to 0} (\sin x)x = 0 \cdot 0 = 0$$ Step 4: Evaluate the limit of the second term as $n \to \infty$: $$\lim_{n\to\infty} \frac{n^2}{n^2+1}$$ Divide the numerator and denominator by $n^2$: $$\lim_{n\to\infty} \frac{\frac{n^2}{n^2}}{\frac{n^2}{n^2}+\frac{1}{n^2}} = \lim_{n\to\infty} \frac{1}{1+\frac{1}{n^2}}$$ As $n \to \infty$, $\frac{1}{n^2} \to 0$. Therefore, the limit is: $$\frac{1}{1+0} = 1$$ Step 5: Combine the limits of the two terms to find $f(0)$: $$f(0) = \lim_{n\to\infty} \left[ (\sin e^{-n^2})e^{-n^2} + \frac{n^2}{n^2+1} \right]$$ $$f(0) = \lim_{n\to\infty} (\sin e^{-n^2})e^{-n^2} + \lim_{n\to\infty} \frac{n^2}{n^2+1}$$ $$f(0) = 0 + 1$$ $$f(0) = 1$$