Q.$\lim_{x\to 4}\dfrac{4x + 3}{x - 2}$

The key idea is that for a rational function where the denominator does not vanish at the limit point, the limit is simply the function's value at that point.

Since $x-2 \neq 0$ when $x=4$, we can directly substitute:

1. Substitute $x = 4$ into the numerator: $4(4) + 3 = 16 + 3 = 19$. …

Why Direct Substitution Works Here

When you see a limit of a rational function (a polynomial divided by a polynomial), your first instinct should be to check what happens to the denominator at the target point. The key idea is simple: a rational function is continuous wherever its denominator is non-zero. Continuity means the limit equals the function value — just plug in the number.

The classic pitfall students fall into is assuming every limit requires factoring or cancellation. That's only needed when the denominator also goes to zero, creating a $\frac{0}{0}$ form. Here, at $x = 4$, the denominator $x - 2$ equals $2$, which is perfectly fine. So we can skip all the algebraic gymnastics.

Step-by-Step Solution

1. Check the denominator.

   At $x = 4$, the denominator is $4 - 2 = 2 \neq 0$. This tells us the function is continuous at $x = 4$, so the limit is simply the function value.

2. Substitute directly. …