Why is $\lim\limits_{x \to 0} \frac{\sin(5x)}{x} = 5$? Is there some trig identity according to which $\sin(cx) = c\cdot\sin(x)$ (or any identity that could help solve this problem)? I already know that $\lim\limits_{x \to 0} \frac{\sin(x)}{x} = 1$, but I'm not sure exactly how to proceed in this particular case. Thanks in advance.
Why is the limit as $x$ approaches $0$ of $\frac{\sin(5x)}{x} = 5$?.
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limits
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1Make the substitution $x\mapsto \frac y5$. – 2017-01-28
3 Answers
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It is well known limit that $\lim _{ x\rightarrow 0 }{ \frac { \sin { x } }{ x } =1 } $ so $$\\ \lim _{ x\rightarrow 0 }{ \frac { \sin { 5x } }{ x } =\lim _{ x\rightarrow 0 }{ 5\cdot \frac { \sin { 5x } }{ 5x } } } =5$$
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Multiply the expression by 5/5, then you have $$\frac{sin5x}{5x}\cdot5$$ which is basically $5\cdot1$
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An alternative approach would be to use L'Hopital's rule.
$$\lim_{x\to 0} \frac{\sin(5x)}{x}=\lim_{x\to 0} \frac{\frac{d}{dx}(\sin(5x))}{\frac{d}{dx}(x)}=\lim_{x\to 0} \frac{5\cos(5x)}{1}=5$$
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1It is already assumed that $\lim\frac{\sin x}x=1$, so it is probably OK, but note that you need to know that limit in order to deduce that the derivative of $\sin$ is $\cos$. I would therefore be careful when using l'Hopital anywhere near that limit. – 2017-01-28