Let {x} denote the fractional part of x and f(x) = (cos⁻¹(1-{x}²) sin⁻¹(1-{x})) / ({x}-{x}³), x ≠ 0.

Q: Wait, why is {x} = x+1 for x < 0? Shouldn't it just be x?

That is the most common trap! Remember, the definition is {x} = x - \lfloor x \rfloor. When x is a tiny negative number (like -0.001), the greatest integer less than or equal to x is -1, not 0. So, {x} = x - (-1) = x + 1. Always check your floor function value when crossing an integer boundary!

Q: Can I just use L'Hopital's Rule here? It seems faster.

While L'Hopital's Rule is a powerful weapon, it can be a double-edged sword here. Differentiating inverse trigonometric functions with nested fractional parts can lead to a messy algebraic nightmare. Using standard limits like \lim_{h \to 0} \frac{\sin^{-1} h}{h} = 1 is much more elegant and less prone to calculation errors.

Q: Why does the \sin^{-1}(1-h) term simplify to \pi/2 in the RHL but not in the LHL?

Great observation! In the RHL, as h \to 0, the argument (1-h) approaches 1, so \sin^{-1}(1) = \pi/2. However, in the LHL, we substituted {x} = 1-h. The term becomes \sin^{-1}(1-(1-h)), which simplifies to \sin^{-1}(h). As h \to 0, this approaches \sin^{-1}(0) = 0. The behavior changes because the argument of the function changes based on the side of the limit.

Question

Let $\{x\}$ denote the fractional part of $x$ and $f(x) = \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, x 
eq 0$. If $L$ and $R$ respectively denotes the left hand limit and the right hand limit of $f(x)$ at $x = 0$, then $\frac{32}{\pi^2} (L^2 + R^2)$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The function given is:
$$f(x) = \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}$$

The fractional part function $\{x\}$ is defined as $x - \lfloor x floor$. As we approach $x = 0$ from the right ($x 	o 0^+$), $\lfloor x floor = 0$, so $\{x\} = x$.

However, as we approach $x = 0$ from the left ($x 	o 0^-$), $\lfloor x floor = -1$. Thus, $\{x\} = x - (-1) = x + 1$. This shift creates a jump discontinuity that we must handle separately for the left and right limits.

### The Right-Hand Limit (RHL)

To evaluate the limit from the right, we substitute $\{x\} = h$, where $h 	o 0^+$. The expression becomes:
$$R = \lim_{h 	o 0} \frac{\cos^{-1}(1-h^2) \sin^{-1}(1-h)}{h - h^3}$$

Factoring the denominator as $h(1-h^2)$, we note that as $h 	o 0$, $\sin^{-1}(1-h) 	o \sin^{-1}(1) = \frac{\pi}{2}$. We can extract this constant:
$$R = \frac{\pi}{2} \lim_{h 	o 0} \frac{\cos^{-1}(1-h^2)}{h}$$

Using the substitution $\cos^{-1}(1-h^2) = 	heta$, we have $1-h^2 = \cos 	heta$, which implies $h^2 = 1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2})$. Thus, $h = \sqrt{2} \sin(\frac{	heta}{2})$. As $h 	o 0$, $	heta 	o 0$, yielding:
$$R = \frac{\pi}{2} \lim_{	heta 	o 0} \frac{	heta}{\sqrt{2} \sin(\frac{	heta}{2})} = \frac{\pi}{2} \cdot \frac{2}{\sqrt{2}} = \frac{\pi}{\sqrt{2}}$$

### The Left-Hand Limit (LHL)

To evaluate the limit from the left, we set $\{x\} = 1-h$, where $h 	o 0^+$. The expression becomes:
$$L = \lim_{h 	o 0} \frac{\cos^{-1}(1-(1-h)^2) \sin^{-1}(1-(1-h))}{(1-h) - (1-h)^3}$$

Simplifying the terms, the numerator becomes $\cos^{-1}(2h-h^2) \cdot \sin^{-1}(h)$. As $h 	o 0$, $\cos^{-1}(2h-h^2) 	o \cos^{-1}(0) = \frac{\pi}{2}$.

The denominator simplifies to $(1-h)(1-(1-h)^2) = (1-h)(2h-h^2) = h(1-h)(2-h)$. Thus:
$$L = \frac{\pi}{2} \lim_{h 	o 0} \frac{\sin^{-1} h}{h(1-h)(2-h)} = \frac{\pi}{2} \cdot \frac{1}{(1)(2)} = \frac{\pi}{4}$$

### Final Calculation

We have determined $R = \frac{\pi}{\sqrt{2}}$ and $L = \frac{\pi}{4}$. We are tasked with finding the value of $\frac{32}{\pi^2} (L^2 + R^2)$.

Squaring our results, we get $L^2 = \frac{\pi^2}{16}$ and $R^2 = \frac{\pi^2}{2}$. Substituting these into the expression:
$$\frac{32}{\pi^2} \left( \frac{\pi^2}{16} + \frac{\pi^2}{2} ight) = 32 \left( \frac{1}{16} + \frac{8}{16} ight)$$

$$= 32 \left( \frac{9}{16} ight) = 18$$

The final result is **18**.

Enter Numerical Value:

Visualized Solution

Understanding the Function $f (x)$

Behavior of Fractional Part ${x}$ near $x = 0$

Setting up Right Hand Limit ( $R$ )

Simplifying the RHL Expression

Trigonometric Substitution for RHL

Evaluating the RHL Limit

Setting up Left Hand Limit ( $L$ )

Simplifying the LHL Numerator

Simplifying the LHL Denominator

Evaluating the LHL Limit

Final Calculation

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Right-Hand Limit (RHL)

The Left-Hand Limit (LHL)

Final Calculation

Similar Questions

The value of $lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x - [ x ] ^{2} ) \cdot s i n ^{- 1} ( x - [ x ] ^{2} )}{x - x ^{3}}$ , where $[x]$ denotes the greatest integer $\leq x$ is :

$lim_{x \to \frac{π}{2}} \frac{1 - t a n ( x /2 )}{1 + t a n ( x /2 )} \frac{[ 1 - s i n x ]}{[ π - 2 x ] ^{3}}$ is

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ is equal to

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18$ , $f^{'} (3) = 0$ and $f^{''} (3) = 4$ . Then $lim_{x \to 1} (lo g_{e} (\frac{f ( 2 + x )}{f ( 3 )}))^{\frac{18}{( x - 1 ) ^{2}}}$ is equal to :

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ equals

$lim_{x \to - \infty} [\frac{x ^{4} s i n ( 1/ x ) + x ^{2}}{( 1 + ∣ x ∣ ^{3} )}] = .........$

$lim_{x \to 0} (\frac{( x + 2 c o s x ) ^{3} + 2 ( x + 2 c o s x ) ^{2} + 3 s i n ( x + 2 c o s x )}{( x + 2 ) ^{3} + 2 ( x + 2 ) ^{2} + 3 s i n ( x + 2 )})^{\frac{100}{x}}$ is equal to ______.

Let $f (x) = \int_{0}^{x} (t + sin (1 - e^{t})) d t, x \in R$ . Then, $lim_{x \to 0} \frac{f ( x )}{x ^{3}}$ is equal to

If $β = lim_{x \to 0} \frac{e ^{x^{3}} - ( 1 - x ^{3} ) ^{1/3} + ( ( 1 - x ^{2} ) ^{1/2} - 1 ) s i n x}{x s i n ^{2} x}$ , then the value of $6 β$ is ______.

If $lim_{x \to 0} \frac{s i n ^{- 1} x - t a n ^{- 1} x}{3 x ^{3}}$ is equal to $L$ , then the value of $(6 L + 1)$ is

Enter Numerical Value:

Visualized Solution

Understanding the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Behavior of Fractional Part .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } {x} near .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Setting up Right Hand Limit ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R)

Simplifying the RHL Expression

Trigonometric Substitution for RHL

Evaluating the RHL Limit

Setting up Left Hand Limit ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L)

Simplifying the LHL Numerator

Simplifying the LHL Denominator

Evaluating the LHL Limit

Final Calculation

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Right-Hand Limit (RHL)

The Left-Hand Limit (LHL)

Final Calculation

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Understanding the Function $f (x)$

Behavior of Fractional Part ${x}$ near $x = 0$

Setting up Right Hand Limit ( $R$ )

Setting up Left Hand Limit ( $L$ )

The value of $lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x - [ x ] ^{2} ) \cdot s i n ^{- 1} ( x - [ x ] ^{2} )}{x - x ^{3}}$ , where $[x]$ denotes the greatest integer $\leq x$ is :

$lim_{x \to \frac{π}{2}} \frac{1 - t a n ( x /2 )}{1 + t a n ( x /2 )} \frac{[ 1 - s i n x ]}{[ π - 2 x ] ^{3}}$ is

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ is equal to

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18$ , $f^{'} (3) = 0$ and $f^{''} (3) = 4$ . Then $lim_{x \to 1} (lo g_{e} (\frac{f ( 2 + x )}{f ( 3 )}))^{\frac{18}{( x - 1 ) ^{2}}}$ is equal to :

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ equals

$lim_{x \to - \infty} [\frac{x ^{4} s i n ( 1/ x ) + x ^{2}}{( 1 + ∣ x ∣ ^{3} )}] = .........$

$lim_{x \to 0} (\frac{( x + 2 c o s x ) ^{3} + 2 ( x + 2 c o s x ) ^{2} + 3 s i n ( x + 2 c o s x )}{( x + 2 ) ^{3} + 2 ( x + 2 ) ^{2} + 3 s i n ( x + 2 )})^{\frac{100}{x}}$ is equal to ______.

Let $f (x) = \int_{0}^{x} (t + sin (1 - e^{t})) d t, x \in R$ . Then, $lim_{x \to 0} \frac{f ( x )}{x ^{3}}$ is equal to

If $β = lim_{x \to 0} \frac{e ^{x^{3}} - ( 1 - x ^{3} ) ^{1/3} + ( ( 1 - x ^{2} ) ^{1/2} - 1 ) s i n x}{x s i n ^{2} x}$ , then the value of $6 β$ is ______.

If $lim_{x \to 0} \frac{s i n ^{- 1} x - t a n ^{- 1} x}{3 x ^{3}}$ is equal to $L$ , then the value of $(6 L + 1)$ is