Skip to content
Exercise 12.1 · Q13

Q.lim⁡x→0sin⁡axbx\lim_{x\to 0}\dfrac{\sin ax}{bx}

Odisha ChseTextbookSubjective· 2mImportance★★★★★est
7% · 13/175 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The limit lim⁡x→0sin⁡axbx\lim_{x\to 0} \frac{\sin ax}{bx} is found by using the standard limit lim⁡t→0sin⁡tt=1\lim_{t\to 0} \frac{\sin t}{t} = 1 and algebraic manipulation. The final value is ab\frac{a}{b}.

The core idea here is the Limit of a Polynomial-like Form — but with a trigonometric twist. You’re not dealing with a polynomial, but the same principle applies: when xx approaches 0, the expression sin⁡axbx\frac{\sin ax}{bx} behaves like a ratio of linear terms because sin⁡ax\sin ax is approximately axax for small xx. The rigorous way to capture this is through the standard limit lim⁡t→0sin⁡tt=1\lim_{t\to 0} \frac{\sin t}{t} = 1.

Why does this work? Because as xx gets very small, axax also gets very small, and the sine function becomes almost indistinguishable from its argument. The limit sin⁡tt→1\frac{\sin t}{t} \to 1 is the mathematical statement of that fact. So we rewrite the given limit to match that form.

  1. Rewrite the expression to isolate sin⁡axax\frac{\sin ax}{ax}.

    Notice that sin⁡axbx=sin⁡axax⋅ab\frac{\sin ax}{bx} = \frac{\sin ax}{ax} \cdot \frac{a}{b}.

    This is valid because sin⁡axbx=1b⋅sin⁡axx=ab⋅sin⁡axax\frac{\sin ax}{bx} = \frac{1}{b} \cdot \frac{\sin ax}{x} = \frac{a}{b} \cdot \frac{\sin ax}{ax}.

  2. Apply the limit to each factor.

    The limit of a product is the product of the limits, provided each limit exists.

    So lim⁡x→0sin⁡axbx=ab⋅lim⁡x→0sin⁡axax\lim_{x\to 0} \frac{\sin ax}{bx} = \frac{a}{b} \cdot \lim_{x\to 0} \frac{\sin ax}{ax}.

  3. Use the substitution t=axt = ax.

    As x→0x \to 0, t→0t \to 0 as well. Therefore, lim⁡x→0sin⁡axax=lim⁡t→0sin⁡tt=1\lim_{x\to 0} \frac{\sin ax}{ax} = \lim_{t\to 0} \frac{\sin t}{t} = 1. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.