Q.$\lim_{x \to 0} \dfrac{(1 + x)^n - 1}{x}$ is
(A) $n$
(B) $1$
(C) $-n$
(D) $0$

The key idea is that this limit matches the standard definition of the derivative of $f(x) = (1+x)^n$ at $x = 0$.

Recall the definition of the derivative:

Here, $f(x) = (1+x)^n$, so $f(0) = 1$. Therefore,

Now differentiate: $f'(x) = n(1+x)^{n-1}$. Evaluating at $x = 0$ gives $f'(0) = n(1)^{n-1} = n$. …

The core idea here is that the expression $\frac{(1+x)^n - 1}{x}$ looks like the slope of a secant line. As $x$ gets very close to $0$, that secant line becomes the tangent line at $x=0$ — and the slope of that tangent is exactly the derivative of $(1+x)^n$ evaluated at $x=0$.

Why does this work? Because the limit of a polynomial as $x \to 0$ can often be found by direct substitution, but here we have a $0/0$ form. That’s a signal: the numerator and denominator both vanish, so the limit depends on how fast each one shrinks. The binomial expansion lets us see exactly what cancels.

Let’s walk through it step by step.

1. Recognise the form.

   If you plug $x = 0$ directly into $\frac{(1 + x)^n - 1}{x}$, you get $\frac{1^n - 1}{0} = \frac{0}{0}$. That’s indeterminate, so we need to simplify.

2. Expand $(1+x)^n$ using the binomial theorem.

   For any real $n$ (not just a positive integer), the binomial expansion near $x=0$ is:

This is an infinite series if $n$ is not a positive integer, but it still works for small $x$.

3. Subtract 1 and divide by $x$.

Now divide every term by $x$:

4. Take the limit as $x \to 0$. …