Q.$\lim_{x\to 0}\dfrac{\cos x}{\pi - x}$

Concept: Direct substitution for continuous functions

The function $f(x) = \dfrac{\cos x}{\pi - x}$ is a quotient of two continuous functions. We check whether the denominator vanishes at $x = 0$:

$\pi - 0 = \pi \neq 0$ …

When you see a limit problem, the first instinct should be to check whether direct substitution is valid. A rational function $\frac{f(x)}{g(x)}$ has a limit at $x = a$ by direct substitution if both $f$ and $g$ are continuous at $a$ and $g(a) \neq 0$. If the denominator vanishes, you get an indeterminate form and need techniques like L'Hôpital's rule or algebraic manipulation. But if the denominator is non-zero, you're done in one step.

Here, the numerator is $\cos x$ (continuous everywhere) and the denominator is $\pi - x$ (a polynomial, also continuous everywhere). At $x = 0$, the denominator becomes $\pi - 0 = \pi \neq 0$. No indeterminate form, no drama.

Solution

1. Check continuity of numerator and denominator.

   Both $\cos x$ and $\pi - x$ are continuous at $x = 0$.

2. Evaluate the denominator at $x = 0$.

   $\pi - 0 = \pi \neq 0$ …