Limit of $\frac{\arccos x}{\sqrt{1-x}}$ as $x \to 1-$

[gop]

Homework Statement

lim_{x\rightarrow1-}\frac{\arccos x}{\sqrt{1-x}}=?

Hint: substitute x=cos(t)

Homework Equations

The Attempt at a Solution

I, as usual, tried to factor the terms and multiplied with the binomial theorems (to get rid of the square root in the denominator). However, this doesn't lead to any two terms that would cancel out.
I also tried to use the squeeze theorem but ended up with a range from zero to infinity (So I had no two functions that have the same limit).

The solution is \sqrt{2} according to my CAS.

 

[Kreizhn]

The problem is that this is an indeterminate limit in it's current form. Thus we must apply L'Hopitals rule. You'll find that the derivative of the top is

\displaystyle - \frac{1}{\sqrt{1-x^2}}

and the bottom is

\displaystyle - \frac{1}{2\sqrt{1-x}}

When you put it all together, collect under a single root, and make all necessary cancellations, you're left with the limit of a continuous function, namely

\displaystyle \lim _{x \to 1} \frac{2}{\sqrt{x+1}}

 

[gop]

Hi

First of all, thanks for your answer. My first guess was also to go with L'Hopitals rule; however, we haven't done differentiation/integration yet and so I doubt that we are allowed to use it.

However, I'm not really sure how do do it then else especially when considering the hint about substituting x with cos(t).

thx

 

[Sourabh N]

you don't need L'hopital's rule. when you put x = cos t you denom will be like
\sqrt{1 - cos t} = \sqrt{2 sin^{2}(t/2)}
= \sqrt{2} sin t/2 , and num is t; now you can put the limit.

 

[gop]

well I use, as mentioned in the previous post, the identities of sine and cosine and I also changed the limit from x->1 to t->0, to get
lim_{t\rightarrow{0-}}\frac{\sqrt{2}\cdot t}{2\cdot\sin t/2}

However, the problem still exists. When I plug in t=0 I get still 0/0.a

 

[Sourabh N]

Actually, the val for this particular limit : t / sin t is found to be 1 and is used as an identity(I mean whenever this form comes, we can directly put the limit value for it as 1).

 

[gop]

thx for your effort now I got it