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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 18 of 72
Marks: +1, -0
lim⁡x→0ax+xcos⁡xbsin⁡x\lim\limits_{x\rightarrow 0}\frac{ax+x\cos x}{b\sin x}
Solution:  
We have, lim⁡x→0ax+xcos⁡xbsin⁡x\lim\limits_{x\rightarrow 0}\frac{ax+x\cos x}{b\sin x}
= lim⁡x→0x(a+cos⁡x)bsin⁡x\lim\limits_{x\rightarrow 0}\frac{x(a+\cos x)}{b\sin x} = lim⁡x→0xsin⁡xa+cos⁡xb\lim\limits_{x\rightarrow 0}\frac{x}{\sin x}\frac{a+\cos x}{b}
= (lim⁡x→0xsin⁡x)⋅(lim⁡x→0a+cos⁡xb)\left(\lim\limits_{x\rightarrow 0}\frac{x}{\sin x}\right)\cdot\left(\lim\limits_{x\rightarrow 0}\frac{a+\cos x}{b}\right) = (1) . a+cos⁡0b\frac{a+\cos 0}{b} = a+1b\frac{a+1}{b}.
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