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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 20 of 72
Marks: +1, -0
lim⁡x→0sin⁡ax+bxax+sin⁡bx\lim\limits_{x\rightarrow 0}\frac{\sin ax+bx}{ax+\sin bx} , a , b , a + b ≠ 0
Solution:  
We have, lim⁡x→0sin⁡ax+bxax+sin⁡bx\lim\limits_{x\rightarrow 0}\frac{\sin ax+bx}{ax+\sin bx}
Dividing the numerator and denominator by x, we get
lim⁡x→0sin⁡axx+ba+sin⁡bxx\lim\limits_{x\rightarrow 0}\frac{\frac{\sin ax}{x}+b}{a+\frac{\sin bx}{x}} = lim⁡x→0sin⁡axax⋅a+ba+sin⁡axbx⋅b\lim\limits_{x\rightarrow 0}\frac{\frac{\sin ax}{ax}\cdot a + b}{a+\frac{\sin ax}{bx}\cdot b} = a[lim⁡x→0sin⁡axax]+ba+[lim⁡x→0sin⁡bxbx]⋅b\frac{a\left[\lim\limits_{x\rightarrow 0}\frac{\sin ax}{ax}\right]+b}{a+\left[\lim\limits_{x\rightarrow 0}\frac{\sin bx}{bx}\right]\cdot b}
= a(1)+ba+b⋅(1)\frac{a(1)+b}{a+b\cdot(1)} = a+ba+b\frac{a+b}{a+b} = 1.
Since lim⁡θ→0sin⁡θθ\lim\limits_{\theta\rightarrow 0}\frac{\sin\theta}{\theta} = 1.
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