Finding the Limit of f(x) at 0

[Saitama]

Homework Statement

Let f(x)=\frac{sin^{-1}(1-\{x\})\cdot cos^{-1}(1-\{x\})}{\sqrt{2\{x\}}\cdot (1-\{x\})} then find \lim_{x→0^+}f(x) and \lim_{x→0^-}f(x), where {x} denotes the fractional part function.

Homework Equations

The Attempt at a Solution

I have solved \lim_{x→0^-}f(x), using \lim_{g(x)→0} \frac{sin^{-1}g(x)}{g(x)}=1. If we approach a fractional part function at 0 from left, we get the value as 1. Therefore i get my answer to be \frac{\pi}{2\sqrt{2}}/

I am stuck for the first part, \lim_{x→0^+}f(x). When we approach the fractional part function at 0 from right, its value becomes zero. Due to this i get a 0/0 form.
I am not allowed to use L'Hôpital's rule.

Any help is appreciated.

 

[Infinitum]

Solving this without L'Hospital's is fun! Here's how I approached it,

Since x\to 0^+, the fractional part of x, i.e \left \{x \right \} will behave as x.

This gives you the equation as,

\frac{sin^{-1}(1-x)\cdot cos^{-1}(1-x)}{\sqrt{2x}\cdot (1-x)}

Separating,

\lim_{x\to 0^+}\frac{sin^{-1}(1-x)}{(1-x)} \cdot \lim_{x\to 0^+} \frac{cos^{-1}(1-x)}{\sqrt{2x}}

The limit of the first part is trival, and comes out to be \pi/2. The second part is the one that is confusing(without L'Hospital's). Can you try it out?

 

[Saitama]

Solving this without L'Hospital's is fun! Here's how I approached it,

Since x\to 0^+, the fractional part of x, i.e \left \{x \right \} will behave as x.

This gives you the equation as,

\frac{sin^{-1}(1-x)\cdot cos^{-1}(1-x)}{\sqrt{2x}\cdot (1-x)}

Separating,

\lim_{x\to 0^+}\frac{sin^{-1}(1-x)}{(1-x)} \cdot \lim_{x\to 0^+} \frac{cos^{-1}(1-x)}{\sqrt{2x}}

The limit of the first part is trival, and comes out to be \pi/2. The second part is the one that is confusing(without L'Hospital's). Can you try it out?

Thanks Infinitum! I too was stuck at the same point.
I have figured it out, i solved the second part and it came out be one.
Here are the steps:
\lim_{x\to 0^+} \frac{cos^{-1}(1-x)}{\sqrt{2x}}=\lim_{x\to 0^+} \frac{sin^{-1}\sqrt{2x-x^2}}{\sqrt{2x}}
=\frac{sin^{-1}(\sqrt{x}\cdot\sqrt{2-x})}{\sqrt{2x}}
=\frac{sin^{-1}(\sqrt{x}\cdot\sqrt{2-x})\cdot\sqrt{2-x}}{\sqrt{2}\cdot\sqrt{x}\cdot\sqrt{2-x}}

Using \lim_{g(x)→0} \frac{sin^{-1}g(x)}{g(x)}=1,
the solution of limit is 1 and hence the answer is \frac{\pi}{2}.

Thanks once again.

 

[Infinitum]

Thanks Infinitum! I too was stuck at the same point.
I have figured it out, i solved the second part and it came out be one.
Here are the steps:
\lim_{x\to 0^+} \frac{cos^{-1}(1-x)}{\sqrt{2x}}=\lim_{x\to 0^+} \frac{sin^{-1}\sqrt{2x-x^2}}{\sqrt{2x}}
=\frac{sin^{-1}(\sqrt{x}\cdot\sqrt{2-x})}{\sqrt{2x}}
=\frac{sin^{-1}(\sqrt{x}\cdot\sqrt{2-x})\cdot\sqrt{2-x}}{\sqrt{2}\cdot\sqrt{x}\cdot\sqrt{2-x}}

Using \lim_{g(x)→0} \frac{sin^{-1}g(x)}{g(x)}=1,
the solution of limit is 1 and hence the answer is \frac{\pi}{2}.

Thanks once again.

Yep! That's correct!

To me, that first step transformation was the most troublesome, glad you figured it out!