Let's solve the given limits one by one:

(a) $\lim_{x \to \pi} \sin(x - \sin x)$

• At $x = \pi$, $\sin x = 0$, so $x - \sin x = \pi - 0 = \pi$.
• Therefore: $\lim_{x \to \pi} \sin(x - \sin x) = \sin(\pi) = 0.$

(b) $\lim_{t \to 0} \sin\left(\frac{\pi}{2} \cos(\tan t)\right)$

• As $t \to 0$, $\tan t \to 0$ and $\cos(\tan t) \to \cos(0) = 1$.
• Substituting, we get: $\sin\left(\frac{\pi}{2} \cos(\tan t)\right) \to \sin\left(\frac{\pi}{2} \cdot 1\right) = \sin\left(\frac{\pi}{2}\right) = 1.$
• Thus, the limit is: $\boxed{1}.$

(c) $\lim_{y \to 1} \sec\left(y \sec^2 y - \tan^2 y - 1\right)$

• Substitute $y = 1$: $\sec^2(1) - \tan^2(1) - 1 = \sec^2(1) - (\sec^2(1) - 1) - 1 = 1.$
• So: $\lim_{y \to 1} \sec\left(y \sec^2 y - \tan^2 y - 1\right) = \sec(1) = 1.$

(d) $\lim_{x \to 0} \tan\left(\frac{\pi}{4} \cos(\sin x^{1/3})\right)$

• As $x \to 0$, $x^{1/3} \to 0$ and $\sin x^{1/3} \to 0$, so $\cos(\sin x^{1/3}) \to \cos(0) = 1$.
• Substituting: $\tan\left(\frac{\pi}{4} \cdot \cos(\sin x^{1/3})\right) \to \tan\left(\frac{\pi}{4} \cdot 1\right) = \tan\left(\frac{\pi}{4}\right) = 1.$
• Thus: $\boxed{1}.$

(e) $\lim_{t \to 0} \cos\left(\frac{\pi}{\sqrt{19} - 3 \sec 2t}\right)$

• As $t \to 0$, $\sec(2t) \to \sec(0) = 1$, so $\sqrt{19} - 3\sec(2t) \to \sqrt{19} - 3$.
• Substituting: $\cos\left(\frac{\pi}{\sqrt{19} - 3 \sec 2t}\right) \to \cos\left(\frac{\pi}{\sqrt{19} - 3}\right).$
• The final value is: $\cos\left(\frac{\pi}{\sqrt{19} - 3}\right).$

(f) $\lim_{x \to \frac{\pi}{6}} \sqrt{\csc^2 x + 5\sqrt{3} \tan x}$

• At $x = \frac{\pi}{6}$, $\csc x = 2$ and $\tan x = \frac{1}{\sqrt{3}}$.
• Substituting: $\csc^2 x = 4, \quad \text{and} \quad 5\sqrt{3} \tan x = 5\sqrt{3} \cdot \frac{1}{\sqrt{3}} = 5.$
• So: $\sqrt{\csc^2 x + 5\sqrt{3} \tan x} = \sqrt{4 + 5} = \sqrt{9} = 3.$

(g) $\lim_{x \to 0^+} \sin\left(\frac{\pi}{2} e^{\sqrt{x}}\right)$

• As $x \to 0^+$, $\sqrt{x} \to 0$, so $e^{\sqrt{x}} \to e^0 = 1$.
• Substituting: $\sin\left(\frac{\pi}{2} e^{\sqrt{x}}\right) \to \sin\left(\frac{\pi}{2} \cdot 1\right) = \sin\left(\frac{\pi}{2}\right) = 1.$

(h) $\lim_{x \to 1} \cos^{-1}(\ln \sqrt{x})$

• At $x = 1$, $\sqrt{x} = 1$ and $\ln(1) = 0$.
• So: $\cos^{-1}(\ln \sqrt{x}) \to \cos^{-1}(0) = \frac{\pi}{2}.$

Let me know which part you'd like more details about!

Related Questions:

1. What are the properties of the $\sec$ function in trigonometric limits?
2. How does the small-angle approximation help in solving trigonometric limits?
3. Why does the limit involving $\cos^{-1}$ converge to $\frac{\pi}{2}$ in part (h)?
4. How does $e^{\sqrt{x}}$ simplify when $x \to 0^+$?
5. What are some key techniques for solving limits involving trigonometric functions?

Tip: When solving limits, always check if substituting directly works first before applying expansions or approximations.