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?

  • $\begingroup$ First of all, you can pull out the $\frac{\sqrt{3}}{2}$. Then, we get $\frac{\sqrt{3}}{2} \lim \limits_{\Delta x \to 0} \frac{\sin{\Delta x}}{\Delta x} = \frac{\sqrt{3}}{2} \cdot 1 = \frac{\sqrt{3}}{2}$. $\endgroup$ – layman Feb 18 '15 at 17:03
  • 1
    $\begingroup$ Looks like a derivative to me... $\endgroup$ – David Mitra Feb 18 '15 at 17:03
  • $\begingroup$ Surely at this point the fact that $\lim \limits_{t\to 0}\left(\frac{\sin (t)}t\right)=1$ is available. $\endgroup$ – Git Gud Feb 18 '15 at 17:03
  • $\begingroup$ 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. $\endgroup$ – nw. Feb 18 '15 at 17:04
  • 1
    $\begingroup$ See this. $\endgroup$ – David Mitra Feb 18 '15 at 17:10

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.