Evaluate $\lim_{x \to 0} \frac{\arccos(\frac{1-x²}{1+x²})}{\arcsin(x)}$

Question

Evaluate: $$\lim_{x \to 0} \frac{\arccos(\frac{1-x²}{1+x²})}{\arcsin(x)}$$

I found this in a calculus book and I just can't seem to get it.

I see that if we substitute $x$ for $\tan(y)$, we get $2y$ in the numerator, but then what do we do with the $\arcsin(\tan(y))$ in the denominator?!

Btw, the answer is $2$ according to the book, and according to WolframAlpha, the both-sided limit does not exist, i.e., its $-2$ from below and $+2$ from above. — Nick Larry, Mar 15 '22 at 10:47
Use $\arcsin x \approx x$ and now the limit is $2y/\tan y$.. — Saturday, Mar 15 '22 at 10:49
Could you share your work after applying $L'hopitals?$ I've edited the title for searchability — tryst with freedom, Mar 15 '22 at 10:49
@Buraian I am not very comfortable with differentiating inverse trigonometric functions, and feel like I must have made a mistake there. Could you try please? — Nick Larry, Mar 15 '22 at 10:52
If you share the working with mistake, we can know you really put in the effort and can give you 'personalized' help. — tryst with freedom, Mar 15 '22 at 10:53
You could use the triangle rule to figure out how to write the arcsin(y) = arcsin(y/sqrt(1-y^2) ) — tryst with freedom, Mar 15 '22 at 10:55

score 3 · Accepted Answer · answered Mar 15 '22 at 10:55

$$\frac{\arccos(\frac{1-x²}{1+x²})}{\arcsin(x)}\overbrace{=}^{\text{substituting }x=\tan y}\frac{2y}{\arcsin(\tan y)}=2\frac{y}{\tan y}\frac{\color{blue}{\tan y}}{\color{blue}{\arcsin(\tan y)}}$$

Since $x=\tan y\to0, y\to 0$ and the blue colored part $\to 1$ and $\frac{y}{\tan y}\to 1$.

Evaluate $\lim_{x \to 0} \frac{\arccos(\frac{1-x²}{1+x²})}{\arcsin(x)}$

1 Answers1