Question:

<div class="solution"> <p><strong>Step 1:</strong> To determine which of the given limits tends to unity, we need to evaluate each limit individually.</p> <p><strong>Step 2:</strong> For option (a), we have \(\displaystyle\lim_{x \to 0} \dfrac{\sin(\tan x)}{\sin x}\). Using the fact that \(\sin x \approx x\) for small \(x\), we can rewrite this limit as \(\displaystyle\lim_{x \to 0} \dfrac{\sin(\tan x)}{\tan x} \cdot \dfrac{\tan x}{\sin x}\). Since \(\tan x \approx x\) for small \(x\), this limit simplifies to \(\displaystyle\lim_{x \to 0} \dfrac{\sin x}{x} \cdot \dfrac{x}{\sin x} = 1\).</p> <p><strong>Step 3:</strong> For option (b), we have \(\displaystyle\lim_{x \to \pi/2} \dfrac{\sin(\cos x)}{\cos x}\). As \(x\) approaches \(\pi/2\), \(\cos x\) approaches \(0\), so we can use the small-angle approximation \(\sin x \approx x\) to rewrite this limit as \(\displaystyle\lim_{x \to \pi/2} \dfrac{\cos x}{\cos x} = 1\).</p> <p><strong>Step 4:</strong> For option (c), we have \(\displaystyle\lim_{x \to 0} \left(\dfrac{1}{x^2} \int_0^x \dfrac{t + t^2}{1 + \sin t}\, dt\right)\). Using the fact that \(\sin t \approx t\) for small \(t\), we can approximate the integrand as \(\dfrac{t + t^2}{1 + t}\). Then, we can evaluate the integral: \(\int_0^x \dfrac{t + t^2}{1 + t}\, dt = \int_0^x \dfrac{t(1 + t)}{1 + t}\, dt = \int_0^x t\, dt = \dfrac{x^2}{2}\). Therefore, the limit becomes \(\displaystyle\lim_{x \to 0} \dfrac{1}{x^2} \cdot \dfrac{x^2}{2} = \dfrac{1}{2}\).</p> <p><strong>Step 5:</strong> For option (d), we have \(\displaystyle\lim_{x \to 0} \dfrac{\displaystyle\int_0^x \sin^2 t\, dt}{\sqrt[3]{1 + x^3} - 1}\). Using the fact that \(\sin^2 t \approx t^2\) for small \(t\), we can approximate the integral as \(\int_0^x t^2\, dt = \dfrac{x^3}{3}\). Then, we can use the fact that \(\sqrt[3]{1 + x^3} - 1 \approx \dfrac{x^3}{3}\) for small \(x\) to rewrite the limit as \(\displaystyle\lim_{x \to 0} \dfrac{\dfrac{x^3}{3}}{\dfrac{x^3}{3}} = 1\).</p> <p><strong>Answer:</strong> The limits in options (a) and (d) tend to unity.</p> <div class="key-concept"><strong>Key Concept:</strong> Small-angle approximations, such as \(\sin x \approx x\) and \(\tan x \approx x\), are useful for evaluating limits involving trigonometric functions. Additionally, using the fundamental theorem of calculus to evaluate definite integrals can help simplify complex limits.</div> </div>