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NCERT Exemplar · Q56

Q.lim⁡x→0(1+x)n−1x\lim_{x \to 0} \dfrac{(1 + x)^n - 1}{x} is
(A) nn
(B) 11
(C) −n-n
(D) 00

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This limit is the definition of the derivative of f(x)=(1+x)nf(x) = (1+x)^n at x=0x=0, which equals nn. The answer is (A).

The core idea here is that the expression (1+x)n−1x\frac{(1+x)^n - 1}{x} looks like the slope of a secant line. As xx gets very close to 00, that secant line becomes the tangent line at x=0x=0 — and the slope of that tangent is exactly the derivative of (1+x)n(1+x)^n evaluated at x=0x=0.

Why does this work? Because the limit of a polynomial as x→0x \to 0 can often be found by direct substitution, but here we have a 0/00/0 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=0x = 0 directly into (1+x)n−1x\frac{(1 + x)^n - 1}{x}, you get 1n−10=00\frac{1^n - 1}{0} = \frac{0}{0}. That’s indeterminate, so we need to simplify.

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

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

(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+⋯(1+x)^n = 1 + n x + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + \cdots

This is an infinite series if nn is not a positive integer, but it still works for small xx.

  1. Subtract 1 and divide by xx.

(1+x)n−1=nx+n(n−1)2x2+n(n−1)(n−2)6x3+⋯(1+x)^n - 1 = n x + \frac{n(n-1)}{2} x^2 + \frac{n(n-1)(n-2)}{6} x^3 + \cdots

Now divide every term by xx:

(1+x)n−1x=n+n(n−1)2x+n(n−1)(n−2)6x2+⋯\frac{(1+x)^n - 1}{x} = n + \frac{n(n-1)}{2} x + \frac{n(n-1)(n-2)}{6} x^2 + \cdots

  1. Take the limit as x→0x \to 0. …
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