Question 5.35

Solutions

leumasicOfficial

3 months ago

We have

\alpha=\lim_{x\to 0}\frac{\sin x}{x}.

i)

\begin{align*} \lim_{x\to\infty}\frac{\sin x}{x} &=\lim_{x\to 0^+}\frac{\sin(1/x)}{1/x} \\ &=\lim_{x\to 0}\frac{\sin(1/x)}{1/x} \\ &=\frac{1}{\alpha}, \end{align*}

assuming $\alpha\neq 0$ .

ii)

\begin{align*} \lim_{x\to\infty}x\sin\left(\frac{1}{x}\right) &=\lim_{x\to 0^+}\frac{1}{x}\sin x \\ &=\lim_{x\to 0}\frac{\sin x}{x} \\ &=\alpha. \end{align*}

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Q 5.35

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Q 5.35