1. **Problem statement:** Find the limits: (i) $$\lim_{x \to \infty} \frac{x \sin x}{x^2 + 4}$$ (ii) $$\lim_{x \to 0} \frac{1 - \cos x}{x^2}$$ (iii) $$\lim_{x \to \infty} \left(\frac{x + 2}{x + 1}\right)^{-x^2}$$ --- 2. **Recall important formulas and rules:** - For large $x$, $\sin x$ oscillates between $-1$ and $1$. - The Taylor expansion near 0: $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$ - Exponential and logarithm limits: $\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x = e^a$ --- 3. **Calculate each limit:** (i) $$\lim_{x \to \infty} \frac{x \sin x}{x^2 + 4} = \lim_{x \to \infty} \frac{\sin x}{x + \frac{4}{x}}$$ Since $\sin x$ is bounded between $-1$ and $1$, and denominator grows without bound, the fraction tends to 0. **Answer:** $$0$$ --- (ii) Use the Taylor expansion of $\cos x$ near 0: $$1 - \cos x = 1 - \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\right) = \frac{x^2}{2} - \frac{x^4}{24} + \cdots$$ Divide by $x^2$: $$\frac{1 - \cos x}{x^2} = \frac{\frac{x^2}{2} - \frac{x^4}{24} + \cdots}{x^2} = \frac{1}{2} - \frac{x^2}{24} + \cdots$$ Taking the limit as $x \to 0$: $$\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$$ --- (iii) Rewrite the expression inside the limit: $$\left(\frac{x + 2}{x + 1}\right)^{-x^2} = \left(1 + \frac{1}{x + 1}\right)^{-x^2}$$ Take the natural logarithm: $$\ln y = -x^2 \ln \left(1 + \frac{1}{x + 1}\right)$$ Use the approximation $\ln(1 + t) \approx t - \frac{t^2}{2}$ for small $t$: $$\ln \left(1 + \frac{1}{x + 1}\right) \approx \frac{1}{x + 1} - \frac{1}{2(x + 1)^2}$$ Multiply by $-x^2$: $$\ln y \approx -x^2 \left(\frac{1}{x + 1} - \frac{1}{2(x + 1)^2}\right) = -\frac{x^2}{x + 1} + \frac{x^2}{2(x + 1)^2}$$ Simplify the dominant term: $$-\frac{x^2}{x + 1} = -x + \cancel{\frac{x^2}{x + 1}}$$ As $x \to \infty$, $\frac{x^2}{x + 1} \sim x$, so: $$\ln y \sim -x + \frac{1}{2}$$ Thus, $$y = e^{\ln y} \to e^{-\infty} = 0$$ --- **Final answers:** (i) $$0$$ (ii) $$\frac{1}{2}$$ (iii) $$0$$