Q.If the function $f(x)$ satisfies $\lim_{x\to 1}\dfrac{f(x) - 2}{x^2 - 1} = \pi$, evaluate $\lim_{x\to 1} f(x)$.

The given limit condition tells us something crucial about the behavior of $f(x)$ near $x = 1$.

Since $\lim_{x\to 1}\dfrac{f(x) - 2}{x^2 - 1} = \pi$ exists and is finite, while the denominator $x^2 - 1 \to 0$ as $x \to 1$, the numerator $f(x) - 2$ must also approach zero. If $f(x) - 2$ approached any nonzero value, the fraction would blow up to infinity, contradicting the finite limit $\pi$ …

When you see a limit of the form $\frac{f(x) - 2}{x^2 - 1}$ that equals a finite number as $x \to 1$, the first instinct should be to ask: what happens to the denominator? Since $x^2 - 1 \to 0$ as $x \to 1$, we're dividing by something that approaches zero. For this quotient to have a finite limit (in this case $\pi$), the numerator must also approach zero—otherwise we'd get an infinite limit or no limit at all.

This is the key insight: a finite limit of a quotient with vanishing denominator forces the numerator to vanish too.

Let me show you why this must be true, then verify it.

Why the numerator must vanish

Suppose $\lim_{x\to 1} f(x) = L$ for some value $L$. Then we can rewrite our given limit as:

If this limit equals $\pi$ (a finite number), and we know that $\lim_{x\to 1}(x^2 - 1) = 0$, then by the product rule for limits:

Therefore $\lim_{x\to 1} f(x) - 2 = 0$, which gives us $\lim_{x\to 1} f(x) = 2$.

Step-by-step verification

1. Identify what we know. We're given that $\lim_{x\to 1}\frac{f(x) - 2}{x^2 - 1} = \pi$, and we need to find $\lim_{x\to 1} f(x)$.

2. Examine the denominator. As $x \to 1$, we have $x^2 - 1 \to 1^2 - 1 = 0$. The denominator vanishes. …