I came across a question involving continuous extension of a trigonometric function: $$ \frac{\sin^2(\frac{1}{x})\sin^3x}{x} $$ if $ x \lt 0$ and $$x(x-1)(x-2)(x-3)(x-4)(x-5)$$ if $x \ge 0 $. How do we go about showing that it is continuous at 0? I know you have to show that $\lim_{x\to 0^-}$ = $\lim_{x\to 0^+}$, but not really sure how to find the answer for $\lim_{x\to 0^-}\frac{\sin^2(\frac{1}{x})\sin^3x}{x}$. Is it possible to explain step by step?
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Use squeeze theorem and $\lim_{x \to 0} \frac {\sin x}x = 1$. – player3236 Nov 15 '20 at 07:48
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Fact 1 $\lim_{x\to0}\frac{\sin x}{x}=1$ See this question
Fact 2 $\lim_{x\to\infty}\frac{\sin x}{x}=0$ (since $|\frac{\sin x}{x}|\le 1/|x|\to0$).
Equivalently, $\lim_{x\to0}x\sin\frac{1}{x}=0$.
Then $$\frac{\sin^2(\frac{1}{x})\sin^3x}{x}=\left(x\sin\frac{1}{x}\right)^2\left(\frac{\sin x}{x}\right)^3\to0^2\times1^3=0$$
Chrystomath
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