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Consider $\lim_{x \to 0} \frac{\sin(\tan x)}{\sin x}$. The answer is $1$. This is clear intuitively since $\tan x ≈ x$ for small $x$. How do you show this rigorously? In general, it does not hold that $\lim_{x \to p} \frac{f(g(x))}{f(x)} = 1$ if $g(x) - x \to 0$ as $x \to p$.

No advanced techniques like series or L'Hôpital. This is an exercise from a section of a textbook which only presumes basic limit laws and continuity of composite continuous functions.

This should be a simple problem but I seem to be stuck. I've tried various methods, including $\epsilon-\delta$, but I'm not getting anywhere. The composition, it seems to me, precludes algebraic simplification.

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  • $\begingroup$ $\tan x \sim x$ means that $\lim_{x \to 0} (\tan x) / x = 1$. $\endgroup$ – Crostul Mar 22 '16 at 22:22
  • $\begingroup$ @Crostul The $ ≈$ is used informally as "approximately equal". $\endgroup$ – MathematicsStudent1122 Mar 22 '16 at 22:23
  • $\begingroup$ Right. So if they are approximately equal then tan x/x is approximately 1 so lim tan/x = 1. $\endgroup$ – fleablood Mar 22 '16 at 22:30
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$$\lim_{x \to 0}\dfrac{\sin(\tan(x))}{\sin(x)}=\lim_{x \to 0}\dfrac{\sin(\tan(x))}{\sin(x)/x} \cdot \dfrac{1}{x} = \lim_{x \to 0}\dfrac{\sin(\tan(x))/\tan(x)}{\sin(x)/x} \cdot \dfrac{\tan(x)}{x}\text{.}$$ Now $$\lim_{x \to 0}\sin(\tan(x))/\tan(x) = \lim_{\tan(x) \to 0}\sin(\tan(x))/\tan(x) = 1\text{,}$$ $$\lim_{x \to 0}\sin(x)/x = 1$$ and you can use this (or any of the other answers if you haven't covered derivatives) to show $$\lim_{x \to 0}\tan(x)/x=\sec(0) = 1\text{.}$$

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$$\frac{\sin(\tan x)}{\sin x} = \frac{\sin(\tan x)}{\tan x} \frac{\tan x}{\sin x} = \frac{\sin(\tan x)}{\tan x}\frac{1}{\cos x}.$$

As $x\to 0, \tan x \to 0,$ hence the first fraction on the right $\to 1.$ We also know $\cos x \to 1,$ so the second fraction on the right $\to 1.$ The limit is therefore $1\cdot 1 = 1$

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Recall that $\tan x = \frac{\sin x}{\cos x}$ and that $\cos x = \sqrt{1 - \sin^2 x}$. Let $u = \sin x$ \begin{align} \lim_{x \to 0} \frac{\sin(\frac{\sin x}{\cos x})}{\sin x} &= \lim_{x \to 0} \frac{\sin(\frac{\sin x}{\sqrt{1 - \sin^2 x}})}{\sin x}\\ &= \lim_{u \to 0} \frac{\sin \frac{u}{\sqrt{1-u^2}}}{\frac{u}{\sqrt{1 - u^2}}} \frac{1}{\sqrt{1 - u^2}}\\ &= 1 \cdot \frac{1}{\sqrt{1 - 0}} = 1 \end{align}

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$$\lim_{x\to 0}\frac{\sin(\tan x)}{\sin x}=\lim_{x\to 0}\frac{\sin(\tan x)}{\sin x}\cdot\frac{\frac{1}{\cos x}}{\frac{1}{\cos x}}=\lim_{x\to 0}\frac{\sec x\sin(\tan x)}{\tan x}$$ Can you take it from here?

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Here we present a solution that relies on only (i) elementary inequalities from geometry and (ii) the squeeze theorem.


NOTE: We first note that $\frac{\sin(\tan (x))}{\sin(x)}$ is an even function of $x$ and hence, if the right-side limit $\lim_{x\to 0^+}\frac{\sin(\tan (x))}{\sin(x)}$ exists, then the limit $\lim_{x\to 0}\frac{\sin(\tan (x))}{\sin(x)}$ exists. The ensuing analysis focuses, therefore, on establishing the right-side limit.


First, recall from elementary geometry that the sine function satisfies the inequalities

$$x\cos(x)\le \sin(x)\le x \tag 1$$

for $0\le x\le \pi/2$. From $(1)$ it is easy to see that

$$x\le \tan(x)\le \frac{x}{\cos(x)} \tag 2$$

for $0\le x<\pi/2$. Using $(1)$ and $(2)$, we can write for $0<x<\arctan(\pi/2)$

$$\cos(\tan(x))\le \frac{\sin(\tan(x))}{\sin(x)}\le \frac{1}{\cos^2(x)} \tag 3$$

Finally, applying the squeeze theorem to $(3)$ yields

$$\lim_{x\to 0^+} \frac{\sin(\tan(x))}{\sin(x)}=1$$

whereupon exploiting the evenness of $\frac{\sin(\tan(x))}{\sin(x)}$ establishes the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0} \frac{\sin(\tan(x))}{\sin(x)}=1}$$

Tools Used:

  • The inequalities in $(1)$ and $(2)$
  • The Squeeze Theorem
  • Continuity of the cosine and tangent functions for $|x|<\pi/2$
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You can always try Taylor expansions.

$$\frac{\sin(\tan x)}{\sin x} = \frac{\tan x - \frac{\tan^3 x}{3!} + \mathcal{O}(\tan x)^5}{\sin x} = \frac{x+\frac{x^3}{3!} + \mathcal{O}(x^5)- \frac{\tan^3 x}{3!} + \mathcal{O}(\tan x)^5}{x - \frac{x^3}{3!} + \mathcal{O}(x^5)} = \frac{1+\frac{x^2}{3!} + \mathcal{O}(x^4)}{1 - \frac{x^2}{3!} + \mathcal{O}(x^4)}- \frac{\frac1x\frac{\tan^3 x}{3!} + \frac1x\mathcal{O}(\tan x)^5}{1 - \frac{x^2}{3!} + \mathcal{O}(x^4)} \rightarrow 1 \text{ as } x\rightarrow 0$$

In the last equality, note that the numerator in the latter term can be expanded as a polynomial with a zero constant term. Thus it vanishes in the limit.

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  • $\begingroup$ Ah... I missed where you wrote "no series." My bad. I'll leave the post up, I guess. $\endgroup$ – zahbaz Mar 23 '16 at 0:42

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