Limits and Derivatives: Common Mistakes and Fixes (5)

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Tags Limits And Derivatives

Question

Evaluate lim⁡x→0tan⁡x−sin⁡xx3\displaystyle\lim_{x \to 0} \tfrac{\tan x - \sin x}{x^3}.

Solution — Step by Step

As x→0x \to 0, both numerator and denominator →0\to 0. So we have 0/00/0 — apply expansion or factoring.

 tan⁡x−sin⁡x=sin⁡x(1cos⁡x−1)=sin⁡x⋅1−cos⁡xcos⁡x\tan x - \sin x = \sin x \left(\tfrac{1}{\cos x} - 1\right) = \sin x \cdot \tfrac{1 - \cos x}{\cos x} lim⁡x→0sin⁡xcos⁡x⋅x3(1−cos⁡x)\lim_{x \to 0} \tfrac{\sin x}{\cos x \cdot x^3}(1 - \cos x) =lim⁡x→0sin⁡xx⋅1−cos⁡xx2⋅1cos⁡x= \lim_{x \to 0} \tfrac{\sin x}{x} \cdot \tfrac{1 - \cos x}{x^2} \cdot \tfrac{1}{\cos x}

lim⁡sin⁡xx=1\lim \tfrac{\sin x}{x} = 1, lim⁡1−cos⁡xx2=12\lim \tfrac{1 - \cos x}{x^2} = \tfrac{1}{2}, lim⁡1cos⁡x=1\lim \tfrac{1}{\cos x} = 1.

lim⁡=1⋅12⋅1=12\lim = 1 \cdot \tfrac{1}{2} \cdot 1 = \tfrac{1}{2}

Final answer: 12\mathbf{\tfrac{1}{2}}.

Why This Works

The standard limits lim⁡x→0sin⁡xx=1\lim_{x \to 0} \tfrac{\sin x}{x} = 1 and lim⁡x→01−cos⁡xx2=12\lim_{x \to 0} \tfrac{1 - \cos x}{x^2} = \tfrac{1}{2} are the building blocks of every “trig limit” question in CBSE and JEE Main. By factoring the original expression into a product of these standard pieces, we sidestep complicated algebra entirely.

The factoring move — tan⁡x−sin⁡x=sin⁡x⋅1−cos⁡xcos⁡x\tan x - \sin x = \sin x \cdot \tfrac{1 - \cos x}{\cos x} — is worth memorising as a pattern: any time you see tan⁡x±sin⁡x\tan x \pm \sin x near zero, try pulling out the sin⁡x\sin x.

Alternative Method

Use Taylor expansions: tan⁡x=x+x3/3+O(x5)\tan x = x + x^3/3 + O(x^5), sin⁡x=x−x3/6+O(x5)\sin x = x - x^3/6 + O(x^5). Subtract:

tan⁡x−sin⁡x=x3(13+16)+O(x5)=x32+O(x5)\tan x - \sin x = x^3\left(\tfrac{1}{3} + \tfrac{1}{6}\right) + O(x^5) = \tfrac{x^3}{2} + O(x^5)

Divide by x3x^3 and take limit: 1/21/2. Same answer, much faster if you remember the series.

Common Mistake

Applying L’Hôpital’s rule three times (since each application gives another 0/00/0). It works in principle but the algebra explodes. The factor-and-recognize approach is far cleaner. JEE expects you to spot the pattern, not grind through derivatives.

When you see a limit with high powers of xx in the denominator (like x3x^3 or x4x^4), think Taylor expansion. The number of leading terms you need from each expansion equals the power in the denominator + 1.

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