Substitution Method

For a trigonometric limit as $x\to a$ where a is a fixed non-zero angle (typically $\pi,\ \pi/2,\ \pi/3,\ \pi/4$ or $\pi/6$), the substitution $x-a=t$ — so $x=a+t$ and, crucially, $t\to0$ as $x\to a$ — converts the problem into a limit at $t=0$, where the standard corollaries of section 7.4 (all stated for an argument tending to zero) become directly usable after expanding $\sin(a\pm t)$, $\cos(a\pm t)$ or $\tan(a\pm t)$ with the angle-sum identities.\n\nWorked pattern 1: $\lim_{x\to\pi/2}\dfrac{\cos x}{x-\pi/2}$: putting $t=x-\pi/2$ gives $\cos(\pi/2+t)=-\sin t$, so the limit becomes $\lim_{t\to0}\dfrac{-\sin t}{t}=-1$.\n\nWorked pattern 2 (a difference-of-cosines needing the sum-to-product identity mid-substitution): $\lim_{x\to a}\dfrac{\cos x-\cos a}{x-a}$: putting $t=x-a$, $\cos(a+t)-\cos a=-2\sin\!\left(a+\tfrac t2\right)\sin\tfrac t2$, so dividing by $t$ and using $\sin(t/2)/t\to1/2$ gives the limit $-\sin a$ — this is the standard 'derivative of cosine' result reached by pure limit algebra, no calculus needed.\n\nWorked pattern 3 (a double angle-shift with a half-angle-squared identity): $\lim_{x\to1}\dfrac{1+\cos\pi x}{(1-x)^2}$: putting $1-x=t$, $\cos(\pi(1-t))=\cos(\pi-\pi t)=-\cos\pi t$, so $1+\cos\pi x=1-\cos\pi t=2\sin^2(\pi t/2)$, and the limit reduces to $2\left(\dfrac{\pi}{2}\right)^2=\dfrac{\pi^2}{2}$.\n\nWorked pattern 4 (angle-sum expansion for tangent, needing the tan-subtraction formula): $\lim_{x\to\pi/3}\dfrac{\sqrt3-\tan x}{\pi-3x}$: putting $t=\pi/3-x$ and expanding $\tan(\pi/3-t)$ with the subtraction formula produces, after simplification …