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I'm trying to help out a friend with Calc 1 and am struggling to find this limit without using l'hopital's or the small angle approximation.

$$\lim_{\Delta x \to 0} \frac{\sin\left(\frac{\pi}{6}+\Delta x\right)-\frac{1}{2}}{\Delta x}$$

Which I can reduce to

$$\lim_{\Delta x \to 0} \frac{\sqrt{3}\cdot\sin{\Delta x}}{2\cdot\Delta x}$$

Which is where I'm stuck. How can I simplify this further without the small angle approx or a taylor series expansion? Is there a way to do it with just trig identities?

Yuriy S
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nw.
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  • First of all, you can pull out the $\frac{\sqrt{3}}{2}$. Then, we get $\frac{\sqrt{3}}{2} \lim \limits_{\Delta x \to 0} \frac{\sin{\Delta x}}{\Delta x} = \frac{\sqrt{3}}{2} \cdot 1 = \frac{\sqrt{3}}{2}$. – layman Feb 18 '15 at 17:03
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    Looks like a derivative to me... – David Mitra Feb 18 '15 at 17:03
  • Surely at this point the fact that $\lim \limits_{t\to 0}\left(\frac{\sin (t)}t\right)=1$ is available. – Git Gud Feb 18 '15 at 17:03
  • It is a derivative, but my friend hasn't learned derivatives yet. I'm trying to do this strictly algebraically, and without any advanced math. – nw. Feb 18 '15 at 17:04
  • There's no way to do this problem without knowing the value of $\lim_{x \rightarrow 0} sin(x)/x$, and there's no way to find that 'strictly algebraically'. Your friend probably already learned this specific limit in class. – Alex Zorn Feb 18 '15 at 17:08
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    See this. – David Mitra Feb 18 '15 at 17:10
  • @nw How did you succeed to reduce the limit to $;\frac{\sqrt3\sin\Delta_x}{2\Delta_x};$ ? For that you'd need to know how to evaluate the limit $$\lim_{\Delta_x\to 0}\frac{\cos\Delta_x-1}{\Delta_x}=0$$ so how do you do it without derivatives, Taylor, l'hospital and stuff? – Timbuc Feb 18 '15 at 18:01
  • @Timbuc I used the identity $\sin{(\phi+\theta)}=\sin{\phi}\cdot\cos{\theta}+\cos{\phi}\cdot\sin{\theta}$ – nw. Feb 18 '15 at 18:04
  • @nw I was sure you did, yet that doesn't answer my question: how did you reduce the limit's expression to what you say you did? – Timbuc Feb 18 '15 at 18:06
  • simply plugging in 0 for $\Delta x$ – nw. Feb 18 '15 at 18:07
  • @Timbuc Hey, to prove the derivatives of sinx and cosx, you need the two basic limits $\lim_\limits{x\to 0}{\frac{\sin x}{x}}$ and $\lim_\limits{x\to 0}{\frac{\cos x - 1}{x}}$, so you cannot say that it is from derivatives... –  Jul 28 '15 at 10:25

1 Answers1

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Note that $\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}$, so we get $\frac{\mathrm{d}}{\mathrm{d}x}\left(\sin(x)\right)$ evaluated at $x=\frac{\pi}{6}$. This is just $\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$.

I think this is the intended method for the question, otherwise I don't think they would have chosen something that looks so similar to a derivative.

Esteemator
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