Question:

Step 1: Evaluate $\displaystyle\lim_{x \to 0} \dfrac{\sin(\tan x)}{\sin x}$. $$ \lim_{x \to 0} \dfrac{\sin(\tan x)}{\sin x} = \lim_{x \to 0} \left( \dfrac{\sin(\tan x)}{\tan x} \cdot \dfrac{\tan x}{x} \cdot \dfrac{x}{\sin x} \right) $$ As $x \to 0$, $\tan x \to 0$. Using the standard limit $\lim_{y \to 0} \dfrac{\sin y}{y} = 1$, we have: $$ \lim_{x \to 0} \dfrac{\sin(\tan x)}{\tan x} = 1 $$ Also, it is known that $\lim_{x \to 0} \dfrac{\tan x}{x} = 1$ and $\lim_{x \to 0} \dfrac{x}{\sin x} = 1$. Therefore, $$ \lim_{x \to 0} \dfrac{\sin(\tan x)}{\sin x} = 1 \cdot 1 \cdot 1 = 1 $$ Step 2: Evaluate $\displaystyle\lim_{x \to \pi/2} \dfrac{\sin(\cos x)}{\cos x}$. Let $u = \cos x$. As $x \to \pi/2$, $u \to \cos(\pi/2) = 0$. The limit becomes: $$ \lim_{u \to 0} \dfrac{\sin u}{u} = 1 $$ Step 3: Evaluate $\displaystyle\lim_{x \to 0} \left(\dfrac{1}{x^2} \int_0^x \dfrac{t + t^2}{1 + \sin t}\, dt\right)$. This limit is of the indeterminate form $\frac{0}{0}$. Applying L'Hôpital's Rule and the Fundamental Theorem of Calculus: $$ \lim_{x \to 0} \dfrac{\int_0^x \dfrac{t + t^2}{1 + \sin t}\, dt}{x^2} = \lim_{x \to 0} \dfrac{\dfrac{x + x^2}{1 + \sin x}}{2x} $$ $$ = \lim_{x \to 0} \dfrac{x(1+x)}{2x(1+\sin x)} $$ For $x \neq 0$, we can cancel $x$: $$ = \lim_{x \to 0} \dfrac{1+x}{2(1+\sin x)} $$ Substituting $x=0$: $$ = \dfrac{1+0}{2(1+\sin 0)} = \dfrac{1}{2(1+0)} = \dfrac{1}{2} $$ This limit does not tend to unity. Step 4: Evaluate $\displaystyle\lim_{x \to 0} \dfrac{\int_0^x \sin^2 t\, dt}{\sqrt[3]{1+x^3}-1}$. This limit is of the indeterminate form $\frac{0}{0}$. Applying L'Hôpital's Rule and the Fundamental Theorem of Calculus: The derivative of the numerator is $\dfrac{d}{dx} \int_0^x \sin^2 t\, dt = \sin^2 x$. The derivative of the denominator is $\dfrac{d}{dx} (\sqrt[3]{1+x^3}-1) = \dfrac{d}{dx} ((1+x^3)^{1/3}-1) = \dfrac{1}{3}(1+x^3)^{-2/3} \cdot (3x^2) = x^2(1+x^3)^{-2/3}$. So the limit becomes: $$ \lim_{x \to 0} \dfrac{\sin^2 x}{x^2(1+x^3)^{-2/3}} $$ This can be rewritten as: $$ \lim_{x \to 0} \left( \left(\dfrac{\sin x}{x}\right)^2 \cdot (1+x^3)^{2/3} \right) $$ Using the standard limit $\lim_{x \to 0} \dfrac{\sin x}{x} = 1$: $$ = \left( \lim_{x \to 0} \dfrac{\sin x}{x} \right)^2 \cdot \lim_{x \to 0} (1+x^3)^{2/3} $$ $$ = (1)^2 \cdot (1+0)^{2/3} = 1 \cdot 1 = 1 $$