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  • <img alt="L=\lim _{x \rightarrow 0} \frac{\sin (\sin x)-\sin x}{a x^5+b x^3+c}=-\frac{1}{12}" src="https://entrancecorner.oncodecogs.com/gif.latex?L%3D%5Clim%20_%7Bx%20%5Crightarrow%200%7D%20%5Cfra

L=\lim _{x \rightarrow 0} \frac{\sin (\sin x)-\sin x}{a x^5+b x^3+c}=-\frac{1}{12}
The value/values of a is

 

Option: 1

\epsilon \ R


Option: 2

2


Option: 3

0


Option: 4

1


Answers (1)

best_answer

We have \lim _{x \rightarrow 0} \frac{\sin (\sin x)-\sin x}{a x^5+b x^3+c}=-\frac{1}{12}
In L.H.S. N^r \rightarrow 0 when x \rightarrow 0

\begin{aligned} & \therefore \mathrm{D}^r \text { must } \rightarrow 0 \\ & \Rightarrow c=0 \end{aligned}

Now, \lim _{x \rightarrow 0} \frac{\sin (\sin x)-\sin x}{a x^5+b x^3}

\begin{aligned} & =\lim _{x \rightarrow 0} \frac{2 \sin \left(\frac{\sin x-x}{2}\right) \cos \left(\frac{\sin x+x}{2}\right)}{a x^5+b x^3} \\ & =\lim _{x \rightarrow 0} \frac{2 \sin \left(\frac{\sin x-x}{2}\right)}{\left(\frac{\sin x-x}{2}\right)} \cdot \frac{\sin x-x}{2} \cdot \frac{\cos \left(\frac{\sin x+x}{2}\right)}{a x^5+b x^3} \\ \end{aligned}\begin{aligned} & =\lim _{x \rightarrow 0}\left(\frac{\sin x-x}{a x^5+b x^3}\right) \\ & =\lim _{x \rightarrow 0} \frac{\cos x-1}{5 a x^4+3 b x^2} \\ & =\lim _{x \rightarrow 0} \frac{-\sin x}{20 a x^3+6 b x} \text { (Using L'Hospital's Rule) } \\ & =\lim _{x \rightarrow 0} \frac{-\sin x}{x} \cdot \frac{1}{20 a x^2+6 b} \\ & =-\frac{1}{6 b} \text { for all } a \in R \end{aligned}

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Ajit Kumar Dubey

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