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Question: Given a real valued function such that, \[f(x)=\left\\{ \begin{aligned} & \dfrac{{{\tan }^{2}}...

Given a real valued function such that,

& \dfrac{{{\tan }^{2}}\left\\{ x \right\\}}{\left( {{x}^{2}}-{{\left[ x \right]}^{2}} \right)},for,x>0 \\\ & 1, \\\ & \\\ & \sqrt{\left\\{ x \right\\}\cot \left\\{ x \right\\}},for,x<0 \\\ \end{aligned} \right.for,x=0$$ Where [x] is the integral part and {x} is the fractional part of x, then 1.$$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=1$$ 2.$$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\cot 1$$ 3.$${{\cot }^{-1}}{{\left( \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x) \right)}^{2}}=1$$ 4.$${{\tan }^{-1}}\left( \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x) \right)=\dfrac{\pi }{4}$$
Explanation

Solution

Hint: Convert Fractional part function to Greatest integer function and solve by substituting x as (0+h) OR (0-h)

Complete step-by-step answer:
If we read the question and options carefully and then we come to know that if we find lim⁡x→0+ f(x)\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x) and lim⁡x→0− f(x)\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x) then we can easily verify all options and choose the correct option/s.
Consider,
\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\tan }^{2}}\left\\{ x \right\\}}{\left( {{x}^{2}}-{{\left[ x \right]}^{2}} \right)}
As we all know that,
x=[x]+xx=[x]+\\{x\\}
∴x=x−[x]\therefore \\{x\\}=x-[x]
Therefore above equation will become,
lim⁡x→0+ f(x)=lim⁡x→0+ tan⁡2(x−[x])(x2−[x]2)\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\tan }^{2}}\left( x-[x] \right)}{\left( {{x}^{2}}-{{\left[ x \right]}^{2}} \right)}
As we have to find the limits of positive side we should substitute,
x as (0+h0+h)
As, x→0+,h→0x\to {{0}^{+}},h\to 0
∴lim⁡x→0+ f(x)=lim⁡h→0 tan⁡2(0+h−[0+h])((0+h)2−[0+h]2)\therefore \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{2}}\left( 0+h-[0+h] \right)}{\left( {{\left( 0+h \right)}^{2}}-{{\left[ 0+h \right]}^{2}} \right)}
As ‘h’ is approaching to ‘0’ therefore it can be considered as fraction,
[0+h]=0\left[ 0+h \right]=0as the value of the greatest integer function is an integer with neglecting the fraction.
∴lim⁡x→0+ f(x)=lim⁡h→0 tan⁡2(h−0)(h2−02)\therefore \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{2}}\left( h-0 \right)}{\left( {{h}^{2}}-{{0}^{2}} \right)}
∴lim⁡x→0+ f(x)=lim⁡h→0 tan⁡2hh2\therefore \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{2}}h}{{{h}^{2}}}
Now we can give Common Square to numerator and denominator,
∴lim⁡x→0+ f(x)=lim⁡h→0 (tan⁡hh)2\therefore \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,{{\left( \dfrac{\tan h}{h} \right)}^{2}}
As we all know that we can give limits inside the bracket,
∴lim⁡x→0+ f(x)=(lim⁡h→0 tan⁡hh)2\therefore \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)={{\left( \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\tan h}{h} \right)}^{2}}
To proceed further we should know the formula given below,
Formula:
lim⁡x→0 tan⁡xx=1\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\tan x}{x}=1
Therefore, we can write above equation as,
∴lim⁡x→0+ f(x)=(1)2\therefore \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)={{\left( 1 \right)}^{2}}
∴lim⁡x→0+ f(x)=1\therefore \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=1………………………………………………… (i)
Also consider,
\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\sqrt{\left\\{ x \right\\}\cot \left\\{ x \right\\}}
As we all know that,
x=[x]+xx=[x]+\\{x\\}
∴x=x−[x]\therefore \\{x\\}=x-[x]
Therefore, we can write above equation as,
∴lim⁡x→0− f(x)=lim⁡x→0− (x−[x])cot⁡(x−[x])\therefore \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\sqrt{\left( x-[x] \right)\cot \left( x-[x] \right)}
As we have to find the limits of negative side we should substitute,
x as (0−h0-h)
As, x→0−,h→0x\to {{0}^{-}},h\to 0
∴lim⁡x→0− f(x)=lim⁡h→0 (0−h−[0−h])cot⁡(0−h−[0−h])\therefore \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\sqrt{\left( 0-h-[0-h] \right)\cot \left( 0-h-[0-h] \right)}
As ‘h’ is approaching to ‘0’ therefore it can be considered as fraction,
As the value of the greatest integer function is an integer with neglecting the fraction.
Value of [0−h]\left[ 0-h \right]lies between -1 to 0 and if we neglect the fraction i.e. h then we are left with -1 and not zero.
∴[0−h]=−1\therefore \left[ 0-h \right]=-1
Now if we substitute the value of [0−h]=−1\left[ 0-h \right]=-1 in above equation we will get,
∴lim⁡x→0− f(x)=lim⁡h→0 (−h−(−1))cot⁡(−h−(−1))\therefore \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\sqrt{\left( -h-(-1) \right)\cot \left( -h-(-1) \right)}
∴lim⁡x→0− f(x)=lim⁡h→0 (−h+1)cot⁡(−h+1)\therefore \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\sqrt{\left( -h+1 \right)\cot \left( -h+1 \right)}
Now we can easily substitute the limits to get the required answer,
∴lim⁡x→0− f(x)=(−0+1)cot⁡(−0+1)\therefore \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\sqrt{\left( -0+1 \right)\cot \left( -0+1 \right)}
∴lim⁡x→0− f(x)=cot⁡1\therefore \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\sqrt{\cot 1}……………………………………………………………………….. (ii)
From (i) and (ii) we can say that option (1) is correct and (2) is incorrect.
Now to check whether the option (3) is correct or note first rewrite it below,
cot⁡−1(lim⁡x→0− f(x))2=1{{\cot }^{-1}}{{\left( \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x) \right)}^{2}}=1
Consider L.H.S.,
L.H.S. (Left Hand Side)=cot⁡−1(lim⁡x→0− f(x))2L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)={{\cot }^{-1}}{{\left( \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x) \right)}^{2}}
If we put the value of equation (ii) in above equation we will get,
L.H.S. (Left Hand Side)=cot⁡−1(cot⁡1)2L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)={{\cot }^{-1}}{{\left( \sqrt{\cot 1} \right)}^{2}}
After squaring we can write it as,
L.H.S. (Left Hand Side)=cot⁡−1(cot⁡1)L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)={{\cot }^{-1}}\left( \cot 1 \right)
Now to proceed further we should know the formula given below,
Formula:
cot⁡−1(cot⁡x)=x{{\cot }^{-1}}\left( \cot x \right)=x
By using the above formula we can write L.H.S. as,
∴L.H.S. (Left Hand Side)=1\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)=1
∴L.H.S. (Left Hand Side)=R.H.S. (Right Hand Side)\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)=R.H.S.\text{ }\left( Right\text{ }Hand\text{ }Side \right)…………………………………… (iii)
Now to check whether the option (4) is correct or note first rewrite it below,
tan⁡−1(lim⁡x→0+ f(x))=π4{{\tan }^{-1}}\left( \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x) \right)=\dfrac{\pi }{4}
∴L.H.S. (Left Hand Side)=tan⁡−1(lim⁡x→0+ f(x))\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)={{\tan }^{-1}}\left( \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x) \right)

Put the value of equation (i) in the above equation,
∴L.H.S. (Left Hand Side)=tan⁡−1(1)\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)={{\tan }^{-1}}\left( 1 \right)
As we know that, tan⁡π4=1\tan \dfrac{\pi }{4}=1 therefore, tan⁡−1(1)=π4{{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}
∴L.H.S. (Left Hand Side)=π4\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)=\dfrac{\pi }{4}
∴L.H.S. (Left Hand Side)=R.H.S. (Right Hand Side)\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)=R.H.S.\text{ }\left( Right\text{ }Hand\text{ }Side \right)……………………………………….. (iv)
From (i), (ii), (iii), and (iv) we can write our answer,
Hence, the answers are Option (1), (3), and (4).

Note: Always try to convert fractional part function to the greatest integer function as it reduces the confusion while converting it to normal form.