How could someone in high-school solve this limit without using L'Hopitals Rule? $$\lim _{x \rightarrow 0} \frac{\sin x \cos x-\sin x}{x^{2}} = 0$$ I assume that it involves the fact that $$ \lim _{x \rightarrow 0} \frac{\sin (x)}{x}=1 $$ but I'm not seeing how.
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1$\begingroup$ Factor out $\frac{\sin(x)}{x}$ and use $1-\cos(x)=2\sin^2(x/2)$ $\endgroup$– Mark ViolaOct 28, 2020 at 18:17
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3 Answers
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Notice that $$\lim_{x\to 0} \frac{\sin x \cos x - \sin x}{x^2} = \lim_{x\to 0} \frac{\sin x}{x} \cdot \frac{\cos x - 1}{x} = \lim_{x\to 0} \frac{\sin x}{x} \cdot \lim_{x\to 0} \frac{\cos x - 1}{x}$$
I'm assuming you already know the derivatives of sine and cosine.
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1$\begingroup$ The second limit can be solved in the following way, provided that we have a proof that $\lim_{x \to 0} \sin x / x = 1$: \begin{align} \lim_{x \to 0} \frac{\cos x - 1}{x}&=\lim_{x \to 0} \frac{(\cos x - 1)(\cos x + 1)}{x(\cos x + 1)} \\ &=\lim_{x \to 0} \frac{\cos^2 x-1}{x(\cos x + 1)} \\ &=\lim_{x \to 0} \frac{-\sin^2 x}{x(\cos x + 1)} \\ &=\lim_{x \to 0} \frac{\sin x}{x} \cdot \lim_{x \to 0} \frac{-\sin x}{\cos x + 1} \\ &=1 \cdot 0 \\ &= 0 \, \text{.}\end{align} Once again, we can avoid using L'Hospital's rule. $\endgroup$– JoeOct 28, 2020 at 18:48
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1$\begingroup$ Alex, why does your approach require that you already know the derivatives of sine and cosine? The first limit can be evaluated using a geometric argument followed by applying the squeeze theorem; and the second limit can be evaluated in the way as shown above. These limits are typically used to prove that the derivative of sine is cosine. $\endgroup$– JoeOct 28, 2020 at 18:54
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$\begingroup$ @Joe My answer does not require using L'Hospital's rule either. Note that the two limits are the definitions of the derivatives of sine and cosine, respectively. The geometric argument you link is simply the main part of the proof that $d/dx \sin x = \cos x$ $\endgroup$– Alex G.Oct 28, 2020 at 19:58
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$\begingroup$ There may have been a misunderstanding. What I was questioning was why you specified 'I'm assuming you already know the derivatives of sine and cosine'—I took that to mean 'this argument requires you to know these derivatives'. Since your answer doesn't require L'Hospital's rule, it is clear that we can use your approach without knowing that the derivative of sine and cosine. And then we can use the fact that $\lim_{x \to 0}\frac{\sin x}{x}=1$ and $\lim_{x \to 0}\frac{\cos x - 1}{x}=0$ to prove that the derivative of sine is cosine. $\endgroup$– JoeOct 28, 2020 at 20:36
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$\begingroup$ My original comment was simply meant to complement your answer—we can use the first limit to evaluate the second. $\endgroup$– JoeOct 28, 2020 at 20:38
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Hint:
$$\sin x\cos x-\sin x=-\sin x(1-\cos x)=-\dfrac{\sin^3x}{1+\cos x}$$
as $(1+\cos x)(1-\cos x)=\sin^2x$
answered Oct 28, 2020 at 18:15
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Hint:we have $$\lim_{x \to 0} \frac{\sin x}{x}.\frac{(-2\sin^2 (x/2))}{x^2/4}.4x=?$$
answered Oct 28, 2020 at 18:16