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  • If           <img alt="\mathrm{f(x)=\frac{\sin 3 x+A \sin 2 x+B \sin x}{x^5}, }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D%5Cfrac%7B

If           \mathrm{f(x)=\frac{\sin 3 x+A \sin 2 x+B \sin x}{x^5}, }
\mathrm{x \neq 0 }   is continuous at x=0, then find A, B and f(0). Do not use series expansion or L Hospital's rule.

Option: 1

1


Option: 2

0


Option: 3

2


Option: 4

5


Answers (1)

best_answer

Using trigonometric identities we have
\mathrm{3-4 \sin ^2 x+2 A \cos x+B =4\left(1-\sin ^2 x\right)+2 A \cos \left(2 \cdot \frac{x}{2}\right)+B-1 }
\mathrm{ =4 \cos ^2\left(2 \cdot \frac{x}{2}\right)+2 A \cos \left(2 \cdot \frac{x}{2}\right)+B-1 }

\mathrm{ =4\left(1-2 \sin ^2 \frac{x}{2}\right)^2+2 A\left(1-2 \sin ^2 \frac{x}{2}\right)+B-1 \\ }

\mathrm{ =16 \sin ^4 \frac{x}{2}-16 \sin ^2 \frac{x}{2}-4 A \sin ^2 \frac{x}{2}+2 A+B+3 \\ }

\mathrm{ =16 \sin ^4 \frac{x}{2}-4(A+4) \sin ^2 \frac{x}{2}+2 A+B+3 }
In order to make the limit finite we must have


\mathrm{ A+4=0 \text { and } 2 A+B+3=0 \quad \Longleftrightarrow \quad A=-4 \text { and } B=5 }
By taking those values we get


\mathrm{ \lim _{x \rightarrow 0} \frac{\sin 3 x-4 \sin 2 x+5 \sin x}{x^5}=\left(\lim _{x \rightarrow 0} \frac{16 \sin ^4 \frac{x}{2}}{x^4}\right)\left(\lim _{x \rightarrow 0} \frac{\sin x}{x}\right) }

\mathrm{ =\left(\lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^4 \\ }

 =1
Since f is continuous at 0 it follows f(0)=1.

Posted by

Ritika Kankaria

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