Find the limit of (ln(1 + 2x) − 2x)/x² as x approaches 0.

Step-by-step solution

1. 1

   Check the form and the order needed. At $x = 0$ the numerator is $\ln 1 - 0 = 0$ and the denominator is $0$, so this is $0/0$. Since the denominator is $x^2$, the numerator has to be expanded at least to order $x^2$ — a first-order expansion would give $0$ and tell you nothing.

2. 2

   Expand ln(1 + u) to second order. The Maclaurin series is

   $\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots$

   Substituting $u = 2x$:

   $\ln(1 + 2x) = 2x - \frac{(2x)^2}{2} + O(x^3) = 2x - 2x^2 + O(x^3)$

   Note $(2x)^2 = 4x^2$, so the quadratic coefficient is $-2$, not $-\tfrac12$.

3. 3

   Subtract 2x — the linear terms cancel by design.

   $\ln(1 + 2x) - 2x = -2x^2 + O(x^3)$

   This cancellation is the whole point of the problem: the $-2x$ is placed there precisely to expose the quadratic behaviour of the logarithm.

4. 4

   Divide by x².

   $\frac{\ln(1 + 2x) - 2x}{x^2} = \frac{-2x^2 + O(x^3)}{x^2} = -2 + O(x)$

   Letting $x \to 0$ kills the $O(x)$ remainder, so the limit is $-2$.

5. 5

   Cross-check with L Hopital rule twice. First application:

   $\lim_{x \to 0} \frac{\frac{2}{1 + 2x} - 2}{2x} = \lim_{x \to 0} \frac{-4x}{2x(1 + 2x)}$

   which simplifies to $\lim_{x \to 0} \frac{-2}{1 + 2x} = -2$. Numerically, at $x = 10^{-4}$ the quotient is $-1.99973$ and at $x = -10^{-4}$ it is $-2.00027$, bracketing $-2$.