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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 13 of 72
Marks: +1, -0
lim⁡x→0sin⁡(ax)bx\lim\limits_{x\to 0}\frac{\sin(ax)}{bx}
Solution:  
We have, lim⁡x→0sin⁡(ax)bx\lim\limits_{x\to 0}\frac{\sin(ax)}{bx} = lim⁡x→0sin⁡(ax)ax⋅axbx\lim\limits_{x\to 0}\frac{\sin(ax)}{ax}\cdot\frac{ax}{bx}
= lim⁡x→0sin⁡(ax)ax⋅ab\lim\limits_{x\to 0}\frac{\sin(ax)}{ax}\cdot\frac{a}{b} = ab(lim⁡x→0sin⁡(ax)ax)\frac{a}{b}\left(\lim\limits_{x\to 0}\frac{\sin(ax)}{ax}\right) = ab\frac{a}{b}
Since lim⁡θ→0sin⁡θ0\lim\limits_{\theta\to 0}\frac{\sin\theta}{0} = 1
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