Evaluate lim(x→0) (tan x - x)/(x - sin x) without L'Hôpital

Question

Evaluate $\displaystyle\lim_{x \to 0} \frac{\tan x - x}{x - \sin x}$ without using L’Hopital’s rule.

(JEE Advanced 2023, similar pattern)

Solution — Step by Step

We use the Maclaurin series expansions around $x = 0$ :

\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots

\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots

\tan x - x = \frac{x^3}{3} + \frac{2x^5}{15} + \cdots

The leading term is $\frac{x^3}{3}$ .

x - \sin x = x - \left(x - \frac{x^3}{6} + \cdots\right) = \frac{x^3}{6} - \frac{x^5}{120} + \cdots

The leading term is $\frac{x^3}{6}$ .

\frac{\tan x - x}{x - \sin x} = \frac{\frac{x^3}{3} + \text{higher order}}{\frac{x^3}{6} + \text{higher order}} = \frac{\frac{x^3}{3}\left(1 + \cdots\right)}{\frac{x^3}{6}\left(1 + \cdots\right)}

As $x \to 0$ , the higher-order terms vanish:

\lim_{x \to 0} \frac{\tan x - x}{x - \sin x} = \frac{1/3}{1/6} = \frac{1}{3} \times \frac{6}{1} = \mathbf{2}

Why This Works

Both numerator and denominator approach 0 as $x \to 0$ (it is a $\frac{0}{0}$ form). The key insight: when two expressions both vanish, the limit depends on the rate at which they vanish. Taylor expansion reveals this rate precisely.

Both $\tan x - x$ and $x - \sin x$ are $O(x^3)$ near zero. The leading coefficients are $1/3$ and $1/6$ respectively, and their ratio gives the limit. Higher-order terms ( $x^5, x^7, \ldots$ ) become negligible compared to $x^3$ as $x \to 0$ .

This technique — comparing leading-order terms — is the most powerful tool for evaluating $0/0$ limits without L’Hopital.

Alternative Method

If the exam allows L’Hopital’s rule, apply it three times (since both numerator and denominator have their first two derivatives also going to 0 at $x = 0$ ):

After 3 applications: $\frac{\text{d}^3/\text{d}x^3 (\tan x - x)}{\text{d}^3/\text{d}x^3 (x - \sin x)} \bigg|_{x=0} = \frac{2}{\cos 0} = \frac{2}{1} = 2$

But Taylor series is cleaner and faster once you know the standard expansions.

For JEE, memorise these leading terms: $\sin x \approx x - x^3/6$ , $\cos x \approx 1 - x^2/2$ , $\tan x \approx x + x^3/3$ , $e^x \approx 1 + x + x^2/2$ , $\ln(1+x) \approx x - x^2/2$ . These six expansions solve 80% of limit problems.

Common Mistake

A frequent error: expanding only to the first term ( $\tan x \approx x$ , $\sin x \approx x$ ) and getting $\frac{x - x}{x - x} = \frac{0}{0}$ . This happens because the first-order terms cancel, so you MUST expand to the cubic term. Always expand one order beyond the cancellation point. If the $x$ terms cancel, go to $x^3$ . If $x^3$ also cancels, go to $x^5$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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