Q.$\lim_{x\to -2}\dfrac{\frac{1}{x} + \frac{1}{2}}{x + 2}$

Concept: Indeterminate form requiring algebraic simplification

The direct substitution gives $\frac{0}{0}$, so we simplify the complex fraction first.

Combine the terms in the numerator over a common denominator:

Substitute back:

$\frac{\frac{2+x}{2x}}{x+2} = \frac{2+x}{2x(x+2)} = \frac{x+2}{2x(x+2)}$ …

This limit initially presents an indeterminate form $\frac{0}{0}$ when we substitute $x = -2$ directly. The numerator $\frac{1}{x} + \frac{1}{2}$ becomes $\frac{1}{-2} + \frac{1}{2} = 0$, and the denominator is also $0$. This signals that both numerator and denominator share a common factor of $(x+2)$, which we need to cancel algebraically before evaluating the limit.

The key insight is to recognize that the complex fraction in the numerator can be simplified by finding a common denominator. Once we do that, the $(x+2)$ factor will reveal itself.

Step-by-step solution:

1. Combine the fractions in the numerator over the common denominator $2x$:

2. Rewrite the entire expression as a division of fractions:

3. Notice that $2 + x = x + 2$, so we can cancel this common factor: $\frac{2+x}{2x(x+2)} = \frac{x+2}{2x(x+2)} = \frac{1}{2x}$ …