Q.$\lim_{x \to \frac{\pi}{4}} \dfrac{\sec^2 x - 2}{\tan x - 1}$ is
(A) $3$
(B) $1$
(C) $0$
(D) $\sqrt{2}$

Concept: Limit of an indeterminate form; direct substitution yields $\frac{0}{0}$, so we factor or use L'Hôpital's rule.

Solution:

Substituting $x = \frac{\pi}{4}$ gives $\sec^2\frac{\pi}{4} = 2$ and $\tan\frac{\pi}{4} = 1$, so both numerator and denominator vanish.

Recall the identity $\sec^2 x = 1 + \tan^2 x$. Substituting:

Factor the numerator as a difference of squares: …

When you substitute $x = \frac{\pi}{4}$ directly, both numerator and denominator vanish: $\sec^2\frac{\pi}{4} = 2$ and $\tan\frac{\pi}{4} = 1$, giving the indeterminate form $\frac{0}{0}$. This signals that numerator and denominator share a common factor that we can cancel.

The key insight is to recognize that $\sec^2 x$ and $\tan x$ are related through the fundamental trigonometric identity $\sec^2 x = 1 + \tan^2 x$. This allows us to rewrite the numerator entirely in terms of $\tan x$, making the common factor visible.

Solution

1. Rewrite the numerator using the Pythagorean identity.

   We know that $\sec^2 x = 1 + \tan^2 x$, so:

2. Factor the numerator as a difference of squares.

   The expression $\tan^2 x - 1$ factors as:

3. Substitute and cancel the common factor.

   The limit becomes:

   $\lim_{x \to \frac{\pi}{4}} \frac{(\tan x - 1)(\tan x + 1)}{\tan x - 1}$ …