Q.Evaluate $\lim_{x \to 0} \dfrac{\sqrt{1 + x^3} - \sqrt{1 - x^3}}{x^2}$.

Concept: Limit Of Polynomial (via rationalisation)

The numerator is a difference of square roots, so we multiply numerator and denominator by the conjugate:

This gives:

The core idea here is that when you see a difference of square roots in a limit that gives $\frac{0}{0}$, your first instinct should be to rationalise the numerator. Why? Because the square roots are hiding a factor of $x^3$ inside them — and rationalising exposes that factor, letting you cancel the $x^2$ in the denominator.

Let’s check the form first. As $x \to 0$:

• $\sqrt{1 + x^3} \to \sqrt{1} = 1$
• $\sqrt{1 - x^3} \to 1$ So numerator goes to $1 - 1 = 0$, denominator goes to $0$. That’s $\frac{0}{0}$, an indeterminate form — we need to do some algebra.

1. Rationalise the numerator Multiply numerator and denominator by the conjugate $\sqrt{1 + x^3} + \sqrt{1 - x^3}$:

The numerator becomes a difference of squares:

So the limit is now:

2. Cancel the common factor We have $x^3$ in the numerator and $x^2$ in the denominator — cancel $x^2$:

Now the denominator no longer goes to $0$ — it goes to $1 + 1 = 2$. The numerator goes to $0$. So the whole thing goes to $0/2 = 0$.

3. Evaluate directly …