Q.$\lim_{x\to 0} x\sec x$

Concept: Limit of a product involving a trigonometric function.

Rewrite the secant in terms of cosine:

As $x \to 0$, the numerator $x \to 0$ and the denominator $\cos x \to \cos 0 = 1$. …

When you see a product involving a trigonometric function at a point where the function is well-behaved, the first instinct should be to check whether direct substitution works. The secant function $\sec x = \frac{1}{\cos x}$ is perfectly defined at $x = 0$ because $\cos 0 = 1 \neq 0$. This means we're not dealing with an indeterminate form, and the limit can be found by straightforward evaluation.

The key insight: as $x \to 0$, one factor ($x$) shrinks to zero while the other ($\sec x$) approaches a finite non-zero value. The product of something vanishing and something bounded must itself vanish.

Step-by-step evaluation

1. Rewrite the expression in terms of cosine.

2. Check continuity at the point of interest.

   The function $\frac{x}{\cos x}$ is continuous at $x = 0$ because:

   • The numerator $x$ is continuous everywhere
   • The denominator $\cos x$ is continuous everywhere and $\cos 0 = 1 \neq 0$ …