# limits of inverse trigonometric functions without L'hospital's rule

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.

• 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") – Clement C. Dec 2 '15 at 15:26
• 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. – marg_ocruz Dec 2 '15 at 15:31
• 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? – Clement C. Dec 2 '15 at 15:33
• My main concern here is to get the limit of the function without using L'Hospital's rule. Just mere manipulation. – marg_ocruz Dec 2 '15 at 15:35

$$\lim_{h\to0}\frac{\cos^{-1}\left(\frac{1}{2}-h\right)-\cos^{-1}\left(\frac{1}{2}\right)}{h}$$ 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. $$=\lim_{\theta\to\frac{\pi}{3}}\frac{\theta-\frac{\pi}{3}}{\cos\left(\frac{\pi}{3}\right)-\cos\theta}$$

$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\theta-\frac{\pi}{3}}{2\sin\left(\frac{\theta}{2}-\frac{\pi}{6}\right)\sin\left(\frac{\theta}{2}+\frac{\pi}{6}\right)}$$

$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\frac{\theta}{2}-\frac{\pi}{6}}{\sin\left(\frac{\theta}{2}-\frac{\pi}{6}\right)\sin\left(\frac{\theta}{2}+\frac{\pi}{6}\right)}$$

$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\frac{\theta}{2}-\frac{\pi}{6}}{\sin\left(\frac{\theta}{2}-\frac{\pi}{6}\right)}\times\lim_{\theta\to\frac{\pi}{3}}\frac{1}{\sin\left(\frac{\theta}{2}+\frac{\pi}{6}\right)}$$

$$=1\times\frac{1}{\frac{\sqrt{3}}{2}}$$

$$=\frac{2}{\sqrt{3}}$$

$$=\frac{2\sqrt{3}}{3}$$

• I never thought of transforming $\cos(\frac{\pi}{3})-\cos(\theta)$ in this manner. Thanks! – marg_ocruz Dec 2 '15 at 15:44
• 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 $\frac{x}{\sin x}$. – Ian Miller Dec 2 '15 at 15:45