Question

Can somebody help me with this inverse trigonometric limit?

Answer 1

Answer:

The limit is infinite

Step-by-step explanation:

L'Hopital's Rule

It's used when we are computing a given rational limit and the result is an indeterminate expression like 0/0. If the limit has is rational function with f(x) in the numerator and g(x) in the denominator, then

$\displaystyle \lim _{x\rightarrow a}\frac{f(x)}{g(x)}=\lim _{x\rightarrow a}\ \frac{f'(x)}{g'(x)}$

We need to compute

$\displaystyle L= \lim _{x\rightarrow 5^-}\ \frac{2\ arccos(x-4)}{x-5}=\frac{2\ arccos\ 1}{5-5}=\frac{0}{0}$

Since the result is an indetermination, we use L'Hopital's rule, by computing f'(x) and g'(x) as follows

$\displaystyle f(x)=2\ arccos(x-4)$

$\displaystyle g(x)=x-5$

Recall the derivative of the arccos function is

$\displaystyle [arccos\ u]'=-\frac{-u'}{\sqrt{1-u^2}}$

Thus:

$\displaystyle f'(x)=[2arccos(x-4)]'=\frac{-2(x-4)'}{\sqrt{1-(x-4)^2}}$

$\displaystyle f'(x)=\frac{-2}{\sqrt{1-(x-4)^2}}$

$\displaystyle g'(x)=1$

Replacing into the original limit, we have

$\displaystyle L=\lim _{x\rightarrow 5^-}\ \frac{-2}{\sqrt{1-(x-4)^2}}$

$\displaystyle L=\frac{-2}{\sqrt{1-1^2}}=-\infty$

$\boxed{\text{The limit is infinite}}$