We have a little manipulation to do in order to be able to use the special limit \[\lim_{x \to 0}\frac{\sin(\text{whatever})}{\text{(the same whatever)}} = 1\] Let’s work to get $3 \theta$ in the denominator, to match the $3 \theta$ in the argument of the $\sin$ in the numerator: \begin{align*} \lim_{\theta \to 0}\dfrac{\sin(3 \theta)}{7 \theta} &= \lim_{\theta \to 0}\dfrac{\sin(3 \theta)}{7\theta} \cdot \frac{3}{3} \\[8px] &= \lim_{\theta \to 0}\dfrac{\sin(3 \theta)}{3\theta} \cdot \frac{3}{7} \\[8px] &= \frac{3}{7} \lim_{\theta \to 0}\dfrac{\sin(3 \theta)}{3\theta} \\[8px] &= \frac{3}{7} \cdot 1 \quad \text{[making use of the special limit]} \\[8px] &= \frac{3}{7} \implies \quad\text{ (B)} \quad \cmark \end{align*}

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The expression reminds us of the special limit \[\frac{1- \cos(x)}{x} = 0,\] but this problem has $\cos(3 \theta)$ in the numerator, and $4 \theta$ in the denominator, so we can’t immediately use the special-limit result. Instead we have a little massaging to do. Let’s view the special limit as \[\frac{1- \cos(\text{whatever})}{\text{(the same whatever)}} = 0,\] and do some multiplying and rearranging as-needed so we use the special limit. Specifically, since our $\cos$ in the numerator has the argument $3 \theta,$ let’s manipulate the epxression so we have $3 \theta$ in the denominator: \begin{align*} \lim_{\theta \to 0}\dfrac{\cos(3 \theta)-1}{4 \theta} &= \lim_{\theta \to 0}\dfrac{\cos(3 \theta)-1}{4 \theta} \cdot \frac{3}{3} \\[8px] &= \lim_{\theta \to 0}\dfrac{\cos(3 \theta)-1}{3 \theta} \cdot \frac{3}{4} \\[8px] &= \frac{3}{4}\lim_{\theta \to 0}\dfrac{\cos(3 \theta)-1}{3 \theta} \end{align*} That looks promising. Let’s now factor a $-1$ out of the numerator to match the special limit perfectly: \begin{align*} \hphantom{\lim_{\theta \to 0}\dfrac{\cos(3 \theta)-1}{4 \theta} } & \hphantom{=\lim_{\theta \to 0}\dfrac{\cos(3 \theta)-1}{3 \theta} \cdot \frac{3}{4} }\\ &= \frac{-3}{4}\lim_{\theta \to 0}\dfrac{1 – \cos(3 \theta)}{3 \theta} \\[8px] &= \left( \frac{-3}{4}\right) \cdot 0 \qquad \text{[now using the special limit]}\\[8px] &= 0 \implies \quad\text{ (C)} \quad \cmark \end{align*}

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We first try Substitution: \[\lim_{x \to 0}\dfrac{\sin(x)}{x^2 – 2x} \overset{?}{=} \frac{0}{0} \] Indeterminate once again. And the original expression doesn’t look immediately like the special limit $\displaystyle{\lim_{x \to 0}\frac{\sin(x)}{x}=1}$. Hmmm.

Well, we see we can factor an x out of the denominator, so let’s start there and see if that gets us anything: \begin{align*} \lim_{x \to 0}\dfrac{\sin(x)}{x^2 – 2x} &= \lim_{x \to 0}\dfrac{\sin(x)}{x(x – 2)} \\[8px] &= \lim_{x \to 0}\dfrac{\sin(x)}{x} \cdot \frac{1}{(x – 2)} \\[8px] &= \left( \lim_{x \to 0}\dfrac{\sin(x)}{x}\right) \left( \lim_{x \to 0}\frac{1}{(x – 2)}\right) \\[8px] &= (1) \left(\frac{-1}{2} \right) \\[8px] &= \frac{-1}{2} \implies \quad\text{ (A)} \quad \cmark \end{align*} Once again the lesson is: even if you can’t immediately see how the solution will play out, take whatever initial step occurs to you and then see where that leads!

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We first try Substitution: \[\lim_{x \to 0}\dfrac{x \big( \sin(x)\big)(1 – \cos(x))}{x^3} \overset{?}{=} \dfrac{0}{0}\] Ah, indeterminate once again. Let’s try massaging and grouping to see if we can make use of the special limits \[\lim_{x \to 0}\frac{\sin(x)}{x}=1 \quad \text{and} \quad \lim_{x \to 0}\frac{1-\cos(x)}{x}=0\] We begin by grouping the terms in a way that makes it easy to identify the special limits: \begin{align*} \lim_{x \to 0}\dfrac{x\big( \sin(x)\big)(1 – \cos(x))}{x^3} &= \lim_{x \to 0} \left(\frac{x}{x}\right)\left(\frac{\sin(x)}{x} \right)\left(\frac{1 – \cos(x)}{x} \right) \\[8px] &= \left(\lim_{x \to 0}\frac{x}{x}\right)\left(\lim_{x \to 0}\frac{\sin(x)}{x} \right)\left(\lim_{x \to 0}\frac{1 – \cos(x)}{x} \right) \\[8px] &= (1)(1)(0) \\[8px] &= 0 \implies \quad\text{ (A)} \quad \cmark \end{align*}

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We first try substitution: $$\frac{\sin 3x}{\sin 5x} = \frac{\sin 0}{\sin 0} = \frac{0}{0} $$ Since this limit is in the form of $\dfrac{0}{0}$, it is indeterminiate — we don’t yet know what it is. We have more work to do. This limit reminds us somewhat of our special limit $$ \lim_{x \to 0} \frac{\sin\text{(whatever)}}{\text{(the same whatever)}} = 1$$ but clearly it isn’t currently in a form where we can use that special limit. So let’s do some manipulation to make it into that form: \begin{align*} \lim_{x \to 0}\frac{\sin 3x}{\sin 5x} &= \lim_{x \to 0}\left( \sin 3x \cdot \frac{3x}{3x} \right) \left(\frac{1}{\sin 5x} \cdot \frac{5x}{5x} \right) \\[8px] &= \lim_{x \to 0}\left(\frac{\sin 3x}{3x} \right) \left(\frac{5x}{\sin 5x} \right) \cdot \frac{3x}{5x} \\[8px] &= \lim_{x \to 0}\left(\frac{\sin 3x}{3x} \right) \left(\frac{5x}{\sin 5x} \right) \cdot \frac{3\cancel{x}}{5\cancel{x}} \\[8px] &= \left(\lim_{x \to 0}\frac{\sin 3x}{3x} \right) \left(\lim_{x \to 0}\frac{5x}{\sin 5x} \right) \cdot \frac{3}{5} \\[8px] &=(1)(1)\cdot \frac{3}{5} \\[8px] &= \frac{3}{5} \implies \quad\text{ (B)} \quad \cmark \end{align*}

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If we first try Substitution, we have \[\lim_{x \to \infty }x \sin \left(\frac{1}{x} \right) = \overbrace{\left[ \lim_{x \to \infty }x\right]}^{\text{this goes to }\infty} \cdot \overbrace{\left[\lim_{x \to \infty } \sin \left(\frac{1}{x} \right)\right]}^{\text{this goes to }0}\] since $\displaystyle{\lim_{x \to \infty}\dfrac{1}{x} = 0}$ and $\sin(0) = 0.$ That means the overall limit, which is the product of $\infty$ and 0, is currently indeterminate: we don’t know what it equals, and so we have more work to do.

This question reminds us somehow of the special limit $\displaystyle{\lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1},$ but is obviously different in several important ways. None the less, let’s try manipulating it to put it in the form of the special limit and see what insights we gain. First, note that $x = \dfrac{1}{\frac{1}{x}},$ and so \begin{align*} \lim_{x \to \infty }x \sin \left(\frac{1}{x} \right) &= \lim_{x \to \infty } \dfrac{1}{\frac{1}{x}} \sin \left(\frac{1}{x} \right) \\[8px] &= \lim_{x \to \infty } \frac{\sin \left(\frac{1}{x} \right)}{\frac{1}{x}} \end{align*} That’s looking closer to our special limit . . . especially if we now recognize again that $\displaystyle{\lim_{x \to \infty}\frac{1}{x} = 0,}$ and so both the argument of the sine function and the denominator here go to 0 as the special limit requires. To make this clearer, let’s make the substitution \[u = \frac{1}{x}\] which means that as $x \to \infty,$ $u \to 0.$ That is, we now have \begin{align*} \lim_{x \to \infty }x \sin \left(\frac{1}{x} \right) &= \lim_{x \to \infty } \frac{\sin \left(\frac{1}{x} \right)}{\frac{1}{x}} \\[8px] &= \lim_{u \to 0}\frac{\sin (u)}{u} \\[8px] &= 1 \implies \text{ (D)} \quad \cmark \end{align*} Once again, algebraic manipulation has let us change the limit we were given into our special limit $\displaystyle{\lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1},$ which gives us the answer we need.

And once again we see that if you don’t know what to do immediately with a limit, an excellent move is to start fiddling with it to rewrite it in whatever way you think you (might) need to in order to be use a special limit, and then see where that takes you. (And keep practicing, since it’s only through practice that you will make “tricks” like this part of your working toolkit.)

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As always, we first try Substitution: $$\lim_{x \to 0}\dfrac{1- \cos(x)}{\sin^2(x)} \overset{?}{=} \frac{1- \cos(0)}{\sin^2(0)} = \frac{1-1}{0} = \frac{0}{0}$$ Yet again we obtain  $\dfrac{0}{0}$,  that indeterminate result: we don’t yet know what the limit is. We have more work to do. Seeing the $\sin^2(x)$ there in the denominator makes us think about the trig identity in the hint, $\sin^2(x) + \cos^2(x) = 1,$ which we can rewrite as $$ \sin^2(x) = 1 – \cos^2(x) $$ Hence \begin{align*} \lim_{x \to 0}\dfrac{1- \cos(x) }{\sin^2(x)} &= \lim_{x \to 0}\frac{1- \cos(x) }{1 – \cos^2(x)} \\[8px] &= \lim_{x \to 0}\frac{1- \cos(x) }{(1 – \cos(x))(1 + \cos(x))} \\[8px] &= \lim_{x \to 0}\frac{\cancel{1- \cos(x)} }{\cancel{(1 – \cos(x))}(1 + \cos(x))} \\[8px] &= \lim_{x \to 0} \frac{1}{1 + \cos(x)} \\[8px] &= \frac{1}{1+\cos(0)} \\[8px] &= \frac{1}{1+1} = \frac{1}{2} \implies \quad\text{ (C)} \quad \cmark \\[8px] \end{align*}

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