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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 14 of 72
Marks: +1, -0
lim⁡x→0sin⁡axsin⁡bx\lim\limits_{x\to 0} \frac{\sin ax}{\sin bx} , a , b ≠ 0
Solution:  
We have, lim⁡x→0sin⁡axsin⁡bx\lim\limits_{x\to 0} \frac{\sin ax}{\sin bx} , a , b ≠ 0
= lim⁡x→0sin⁡axax⋅bxsin⁡bx⋅axbx\lim\limits_{x\to 0} \frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{ax}{bx} = lim⁡x→0sin⁡axax\lim\limits_{x\to 0} \frac{\sin ax}{ax} . lim⁡x→01sin⁡bxbx⋅ab\lim\limits_{x\to 0} \frac{1}{\frac{\sin bx}{bx}} \cdot \frac{a}{b}
= ab\frac{a}{b} . (1) . (1) = ab\frac{a}{b}.
Since lim⁡θ→0sin⁡θθ\lim\limits_{\theta\to 0} \frac{\sin\theta}{\theta} = 1.
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