Limits and Derivatives: Conceptual Doubts Cleared (4)

easy • 2 min read
Tags Limits And Derivatives

Question

Evaluate lim⁡x→0sin⁡5xtan⁡3x\displaystyle\lim_{x \to 0} \frac{\sin 5x}{\tan 3x}.

Solution — Step by Step

As x→0x \to 0, both sin⁡5x→0\sin 5x \to 0 and tan⁡3x→0\tan 3x \to 0. Indeterminate form 0/00/0. We can use the standard limits:

lim⁡x→0sin⁡xx=1,lim⁡x→0tan⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1, \quad \lim_{x \to 0} \frac{\tan x}{x} = 1

Multiply numerator and denominator by xx (or rewrite to expose the standard form):

sin⁡5xtan⁡3x=sin⁡5x5x⋅3xtan⁡3x⋅5x3x=sin⁡5x5x⋅3xtan⁡3x⋅53\frac{\sin 5x}{\tan 3x} = \frac{\sin 5x}{5x} \cdot \frac{3x}{\tan 3x} \cdot \frac{5x}{3x} = \frac{\sin 5x}{5x} \cdot \frac{3x}{\tan 3x} \cdot \frac{5}{3}

As x→0x \to 0:

  • sin⁡5x5x→1\dfrac{\sin 5x}{5x} \to 1
  • 3xtan⁡3x→1\dfrac{3x}{\tan 3x} \to 1
  • 53\dfrac{5}{3} is constant

lim⁡x→0sin⁡5xtan⁡3x=1⋅1⋅53=53\lim_{x \to 0} \frac{\sin 5x}{\tan 3x} = 1 \cdot 1 \cdot \frac{5}{3} = \frac{5}{3}

Final answer: 53\dfrac{5}{3}

Why This Works

The standard limits lim⁡sin⁡x/x=1\lim \sin x/x = 1 and lim⁡tan⁡x/x=1\lim \tan x/x = 1 hold only when the argument inside and the denominator both go to 0 together. So we manipulate to make the argument inside (5x) match the denominator (5x), bring in the missing pieces as constants.

This “make the standard limit appear” technique works for almost every trigonometric limit at 0.

Alternative Method

Use L’Hôpital’s rule (Class 12 / JEE):

lim⁡x→0sin⁡5xtan⁡3x=lim⁡x→05cos⁡5x3sec⁡23x=53\lim_{x \to 0} \frac{\sin 5x}{\tan 3x} = \lim_{x \to 0} \frac{5\cos 5x}{3\sec^2 3x} = \frac{5}{3}

Faster if you have L’Hôpital available, but the standard-limit approach is required at Class 11 level.

Common Mistake

Students write lim⁡(sin⁡5x)/x=5\lim (\sin 5x)/x = 5 but lim⁡(sin⁡5x)/(3x)=1\lim (\sin 5x)/(3x) = 1. The second is wrong. The standard limit needs the argument and the denominator to match. So lim⁡(sin⁡5x)/(3x)=(5/3)lim⁡(sin⁡5x)/(5x)=5/3\lim (\sin 5x)/(3x) = (5/3) \lim (\sin 5x)/(5x) = 5/3.

Quick rule: lim⁡x→0(sin⁡ax)/(sin⁡bx)=a/b\lim_{x \to 0} (\sin ax)/(\sin bx) = a/b and lim⁡x→0(tan⁡ax)/(tan⁡bx)=a/b\lim_{x \to 0} (\tan ax)/(\tan bx) = a/b. Same for any combination of sin⁡\sin and tan⁡\tan. Memorise — appears in 2-3 JEE Main questions per paper.

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