How do I solve this without using L'Hospital's rule? $$\lim_{h\rightarrow0} \frac{\cos^{-1}(\frac{1}{2}-h) -\cos^{-1}(\frac{1}{2})}{h}$$

I already tried letting $\theta=\cos^{-1}(\frac{1}{2}-h)$ gets $\cos\theta=\frac{1}{2}-h$ then $h=\frac{1}{2}-\cos\theta$, replaced all $h$ with this and I'm lost. I think this doesn't help. I'm getting a zero as value or indefinite one which shouldn't be because the value must be $\frac{2\sqrt{3}}{3}$. Please help.

  • $\begingroup$ Since you mention L'Hopital, I assume you are familiar with derivatives, in particular that of $\cos^{-1}$. But then, looking at the expression you wrote... can you recognize something (very) related? (or even more than just "related") $\endgroup$ – Clement C. Dec 2 '15 at 15:26
  • $\begingroup$ unfortunately, i can't show it using derivatives but I know this is somehow similar to the definition of a derivative of a function. i just don't know how to manipulate this without L'Hospital's Rule. $\endgroup$ – marg_ocruz Dec 2 '15 at 15:31
  • $\begingroup$ I am slightly confused about how you would apply L'Hopital anyway without differentiating -- but that may be just me. So your question essentially boils down to "how to compute the derivative of $\cos^{-1}$," is that right? $\endgroup$ – Clement C. Dec 2 '15 at 15:33
  • $\begingroup$ My main concern here is to get the limit of the function without using L'Hospital's rule. Just mere manipulation. $\endgroup$ – marg_ocruz Dec 2 '15 at 15:35

Your start is good.

$$\lim_{h\to0}\frac{\cos^{-1}\left(\frac{1}{2}-h\right)-\cos^{-1}\left(\frac{1}{2}\right)}{h}$$ Let $\theta=\cos^{-1}\left(\frac{1}{2}-h\right)$ so $h=\frac{1}{2}-\cos h$

$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\theta-\frac{\pi}{3}}{\frac{1}{2}-\cos\theta}$$ Rewrite fraction as a trig value and apply sums to products. $$=\lim_{\theta\to\frac{\pi}{3}}\frac{\theta-\frac{\pi}{3}}{\cos\left(\frac{\pi}{3}\right)-\cos\theta}$$







  • $\begingroup$ I never thought of transforming $\cos(\frac{\pi}{3})-\cos(\theta)$ in this manner. Thanks! $\endgroup$ – marg_ocruz Dec 2 '15 at 15:44
  • $\begingroup$ In these sort of problems you always need to work it back to one of the basic limits that we know. In this case I have linear expression over trigonometric so I aimed for $\frac{x}{\sin x}$. $\endgroup$ – Ian Miller Dec 2 '15 at 15:45

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