Question

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Let f(x) = x - [x], (where [x] denotes the greatest integer ≤ x) and
$\sf{g(x) = \lim _{x \to 0} \: \dfrac{ \{f(x) \}^{2n } - 1 }{ \{f(x) \}^{2n} + 1 } }$
then g(x) is equal to?

Options:
A) 0
B) 1
C) -1
D) None of these


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Answer 1

Appropriate Question :-

Let f(x) = x - [x], (where [x] denotes the greatest integer ≤ x) and

$\sf{g(x) = \lim _{n \to \infty } \: \dfrac{ \{f(x) \}^{2n } - 1 }{ \{f(x) \}^{2n} + 1 } }$

then g(x) is equal to?

Options:

A) 0

B) 1

C) -1

D) None of these

$\large\underline{\sf{Solution-}}$

Given that,

f(x) = x - [x], (where [x] denotes the greatest integer ≤ x)

We know,

$\rm :\longmapsto\:x - [x] \: = \: \{x \}$

where,

$\rm :\longmapsto\:\{x \} \in \: [0,1)$

We know,

$\rm :\longmapsto\:x \in \: [0,1) \: then \: {x}^{n} = 0 \: as \: n \: \to \: \infty$

$\bf\implies \:\: \: \{ {x} \}^{2n} = 0 \: as \: n \: \to \: \infty$

So,

Now, Consider

$\rm :\longmapsto\:\sf{g(x) = \lim _{n \to \infty } \: \dfrac{ \{f(x) \}^{2n } - 1 }{ \{f(x) \}^{2n} + 1 } }$

$\rm :\longmapsto\:\sf{g(x) = \lim _{n \to \infty } \: \dfrac{ \{x\}^{2n } - 1 }{ \{x \}^{2n} + 1 } }$

$\rm :\longmapsto\:\sf{g(x) = \: \dfrac{ 0 - 1 }{ 0 + 1 } }$

$\rm :\longmapsto\:\sf{g(x) = \: \dfrac{ - 1 }{ 1 } }$

$\bf\implies \:\:\bf{g(x) = - \: 1}$

Hence, Option (c) is correct

Additional Information :-

$\boxed{ \bf{ \: \lim _{x \to 0} \: \frac{sinx}{x} = 1}}$

$\boxed{ \bf{ \: \lim _{x \to 0} \: \frac{tanx}{x} = 1}}$

$\boxed{ \bf{ \: \lim _{x \to 0} \: \frac{tan {}^{ - 1} x}{x} = 1}}$

$\boxed{ \bf{ \: \lim _{x \to 0} \: \frac{sin {}^{ - 1} x}{x} = 1}}$

$\boxed{ \bf{ \: \lim _{x \to 0} \: \frac{log(1 + x)}{x} = 1}}$

$\boxed{ \bf{ \: \lim _{x \to 0} \: \frac{ {e}^{x} - 1 }{x} = 1}}$

Answer 2

Correct Question :-

Let f(x) = x - [x], where [x] denotes the greatest integer ≤ x.

$\sf{g(x)=\displaystyle \lim_{n\to \infty}\dfrac{\{f(x)\}^{2n}-1}{\{f(x)\}^{2n}+1}}$

To Find :-

Value of g(x).

Solution :-

f(x) = x - [x] = {x}, where {.} represents fractional part function.

Range of fractional part function : [0, 1)

Hence, value of f(x) will be 0 else it will lie between 0 and 1.

Now,

$\sf{If\:n\to\infty\:then\:2n\to \infty}$

So,

$\sf{\{f(x)\}^{2n}=0}$

$\sf{(\because f(x)\in[0,1))}$

Solving g(x),

$\sf{g(x)=\displaystyle \lim_{n\to \infty}\dfrac{\{f(x)\}^{2n}-1}{\{f(x)\}^{2n}+1}}$

Applying limits,

$\sf{g(x)=\dfrac{\{f(x)\}^{\infty}-1}{\{f(x)\}^{\infty}+1}}$

$\sf{g(x)=\dfrac{0-1}{0+1}}$

$\sf{g(x)=\dfrac{-1}{1}}$

$\boxed{\sf{g(x)=-1}}$

Answer :-

Value of g(x) = -1. Hence, option C is correct option.

Extra Note :

Whenever infinity is in power of any proper fraction (numerator < denominator) then it equals to 0.

For example,

$\sf{\left(\dfrac{1}{2}\right)^{\infty}=\dfrac{1}{2^\infty}=\dfrac{1}{\infty}=0}$

$\sf{\left(\dfrac{3}{5}\right)^{\infty}=\dfrac{1}{(1.67)^\infty}=\dfrac{1}{\infty}=0}$