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!