# Explaining how $\lim_{x \to 1^-}\tan(\frac{\pi x}2)=\lim_{x\to\pi^-}\tan(\frac{x}2)=\lim_{x \to {\frac{\pi}{2}}^{-}} \frac{\sin x}{\cos x}$

$$\lim_{x \to 1^{-}} \tan\left(\frac{\pi x}{2}\right)=\lim_{x \to \pi^{-}} \tan\left(\frac{x}{2}\right)=\lim_{x \to {\frac{\pi}{2}}^{-}} \frac{\sin x}{\cos x}$$

How does one get to the second and third lines from the first?

As $$x\to 1$$ from below, $$\pi x \to \pi$$ from below. I will use a change of variable to make it clearer: $$\lim_{x\to1^-}\tan(\frac{\pi x}{2})=\lim_{x\to1^-}\tan(\frac{u}{2}),u=\pi x,=\lim_{u\to\pi}\tan(\frac{u}{2})$$
For the second, we know $$\tan(x)=\frac{\sin(x)}{\cos(x)}$$, and as $$x\to\pi$$ from below, $$x/2\to\pi/2$$ from below. With the same change of variable idea:
$$\lim_{x\to\pi^-}\tan(\frac{x}{2})=\lim_{x\to\pi^-}\frac{\sin(u)}{\cos(u)},u=x/2,=\lim_{u\to\pi^/2^-}\frac{\sin(u)}{\cos(u)}$$
Limits examine quantities or functions as a certain value approaches something. It is completely legitimate to replace what's under the limit sign, so long as you adjust your expression accordingly and make sure the substitution still examines the same limit. E.G: $$\lim_{x\to c}yx\equiv\lim_{yx\to yc}yx\equiv\lim_{u\to yc}u$$. Indeed, you could even simplify the limit further by letting $$v=yc$$ and saying $$\lim_{u\to v}u$$. You do have to be very careful with functions and what affects them, maybe thinking about range and domain and so on, but if you keep the behaviour the same, the limit the same, then changing what $$x$$ is "going to" is fine, so long as you change the impact of $$x$$ on the expression accordingly.
• "It is always completely legitimate to replace what's under the limit sign" is not accurate. For the substitution $\lim_{x\to a}f(g(x)) = \lim_{y\to b} f(y)$ to work, we need assumptions that: 1. Limit on the RHS exists. 2. $\lim_{x\to a} g(x) = b.$ 3. $g$ doesn't obtain value $b$ on $(a-\varepsilon, a+ \varepsilon)\setminus \{a\}$ for some $\varepsilon > 0$. Note that continuity is not assumed on $f$ or $g$, however, in practice, you most likely have that $g$ is continuous and injective. This makes conditions 2. and 3. automatically true, which we do have in this case. Jul 15, 2021 at 10:58