If the function $f(x)=\frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}, x \in \mathbf{R}$, is continuous at $x=0$, then $f(0)$ is equal to :
Solution
<p>$$\begin{aligned}
& \lim _\limits{x \rightarrow 0} f(x)=f(0) \quad \text { (continuous at } x=0) \\
& \lim _{x \rightarrow 0} \frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}
\end{aligned}$$</p>
<p>For limit to exist $\beta=0$</p>
<p>$$\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x+\alpha \sin x}{x^3} \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{(3+\alpha) \sin x-4 \sin ^3 x}{x^3}
\end{aligned}$$</p>
<p>For limit to exist $\alpha+3=0 \Rightarrow \alpha=-3$</p>
<p>$\Rightarrow \lim _\limits{x \rightarrow 0} \frac{-4 \sin ^3 x}{x^3}=-4=f(0)$</p>
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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