Limit of a Trigonometric Function

Standard trigonometric substitution limits. Because $\sin x$ and $\cos x$ are continuous everywhere, $\lim_{x\to a}\sin x=\sin a$ and $\lim_{x\to a}\cos x=\cos a$ always hold by direct substitution — these two facts, combined with ordinary trigonometric identities, are the starting point for every trig-limit calculation, whether it is finally settled by factorization, rationalization or straightforward algebraic simplification.\n\nThe Squeeze (Sandwich) theorem. If $f(x)\le g(x)\le h(x)$ for every x in some open interval around a, and $\lim_{x\to a}f(x)=\lim_{x\to a}h(x)=L$, then $g(x)$ is trapped between two functions both converging to L, so $\lim_{x\to a}g(x)=L$ too — the outer two limits 'squeeze' the middle one to the same value. Worked illustration: given $3x^2+2\le f(x)\le5x^2-6$ for all real x, taking the limit of each bound as $x\to-2$ gives $14\le\lim_{x\to-2}f(x)\le14$, forcing $\lim_{x\to-2}f(x)=14$ exactly.\n\n7.4.2 Theorem: $\lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1$ (θ in radians). Proof (for θ tending to 0 through positive values, $0<\theta<\pi/2$): draw a circle of radius r centred at O, with A on the circle on the positive X-axis and P on the circle so that $\angle AOP=\theta$; drop $PM\perp OX$ so $PM=r\sin\theta$, and extend OP to meet the vertical line through A at B, so $AB=r\tan\theta$. Comparing areas, $\text{Area}(\triangle OAP)<\text{Area(sector }OAP)<\text{Area}(\triangle OAB)$ gives $\tfrac12r^2\sin\theta<\tfrac12r^2\theta<\tfrac12r^2\tan\theta$. Dividing throughout by $\tfrac12r^2\sin\theta$ gives $1<\dfrac{\theta}{\sin\theta}<\dfrac{1}{\cos\theta}$, i.e. $\cos\theta<\dfrac{\sin\theta}{\theta}<1$. Taking the limit as $\theta\to0^+$ and applying the Squeeze theorem (since $\cos\theta\to1$ and the constant bound is already 1) forces $\lim_{\theta\to0^+}\dfrac{\sin\theta}{\theta}=1$. For θ tending to 0 through negative values, writing $\theta=-\phi$ with $\phi\to0^+$ gives $\dfrac{\sin\theta}{\theta}=\dfrac{\sin(-\phi)}{-\phi}=\dfrac{\sin\phi}{\phi}\to1$ as well, since sine is an odd function. Both one-sided limits agree at 1, so the two-sided limit $\lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1$ is established. Seven immediate corollaries follow (listed in the table note above): the reciprocal $\theta/\sin\theta\to1$, the tangent versions $\tan\theta/\theta\to1$ and $\theta/\tan\theta\to1$, and the p-scaled forms $\sin(p\theta)/(p\theta)\to1$, $\tan(p\theta)/(p\theta)\to1$, $p\theta/\sin(p\theta)\to1$, $p\theta/\tan(p\theta)\to1$ for any non-zero constant p.\n\nWorked technique gallery. A recurring first move is to divide numerator and denominator by x (or by whatever variable the sine/tangent argument uses) so that each piece separately matches a corollary above: e.g. $\lim_{x\to0}\dfrac{\sin8x}{\tan4x}=\dfrac{\lim(\sin8x/x)}{\lim(\tan4x/x)}=\dfrac{8}{4}=2$. A second recurring move rewrites $1-\cos x$ as $2\sin^2(x/2)$ (from the half-angle identity $1-\cos A=2\sin^2(A/2)$) whenever a bare $1-\cos(\cdot)$ appears, turning it into a squared sine-over-argument form — used for $\lim_{x\to0}\dfrac{2\sin x-\sin2x}{x^3}=1$ (factoring out $2\sin x(1-\cos x)$ from the numerator first) and for $\lim_{x\to0}\dfrac{\sin x^2(1-\cos x^2)}{x^6}=\dfrac12$ (treating $\theta=x^2\to0$ throughout). A third m …