Question

Which of the following is/ are correct?
(a). $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ x+|x| \right]}{x}=0$,
Where, [x] denotes the greatest integer functions.
(b). \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x{{e}^{\dfrac{1}{x}}}}{1+{{e}^{\dfrac{1}{x}}}}=0\]
(c). \[\underset{x\to 3}{\mathop{\lim }}\,{{(x-3)}^{\dfrac{1}{5}}} sgn (x-3)=0\],
Where, sgn stands for signum function.
(d). \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}|x|}{x}=0\]

Answer 1

Hint: To solve the above problem we have to simply use the properties of functions included in problems and then we have to put the limits, we will get the answer.

Complete step-by-step solution -
We have to check which option is/ are correct we should solve them one by one,
Consider, Option 1
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ x+|x| \right]}{x}=0$
As there is modulus function inserted in the problem we should verify whether the limit exist at x=0.And to verify that we should know the basic definition of modulus function which is given below,
Concept:
If f(x) =|x| then,
\[f(x)=-x\] \[at,x<0\]
\[f(x)=0\] \[at,x=0\]
\[f(x)=x\] \[at,x>0\]
By using the above concept we can write the left hand side of option 1 as shown below,
\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ x+(-x) \right]}{x}\] \[at,x<0\]
\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ x+x \right]}{x}\] \[at,x>0\]
Now consider,
L.H.L. ( Left Hand Limit)\[=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left[ x+(-x) \right]}{x}\]
L.H.L. ( Left Hand Limit)\[=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left[ 0 \right]}{x}\]
As we all know that the greatest integer function of ‘0’ is ‘0’, therefore we can write above equation as,
L.H.L. ( Left Hand Limit)\[=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{0}{x}\]
As the limit of ‘x’ is tending to zero and not equal to zero therefore the value of L.H.L. will become zero.
L.H.L. ( Left Hand Limit)\[=0\]………………………………… (i)
Also,
R.H.L. ( Right Hand Limit)\[=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left[ x+x \right]}{x}\]
R.H.L. ( Right Hand Limit)\[=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left[ 2x \right]}{x}\]
As ‘x’ is tending to zero in positive direction therefore,
Put, \[x=0+h\], As \[x\to {{0}^{+}},h\to 0\]
R.H.L. ( Right Hand Limit)\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left[ 2(0+h) \right]}{(0+h)}\]
As the value of ‘h’ is very much small but is greater than zero therefore ‘h’ lies between zero and one. And therefore the value of h in the greatest integer function will become Zero.
R.H.L. ( Right Hand Limit)\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{0}{h}\]
As ‘h’ is tending to zero and not equal to zero therefore the limit will become zero.
R.H.L. ( Right Hand Limit)\[=0\]…………………………………..(ii)
From (i) and (ii) we can say that,
R.H.L. ( Right Hand Limit)= L.H.L. ( Left Hand Limit)
Therefore option (1) is correct.
Consider Option (2),
\[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x{{e}^{\dfrac{1}{x}}}}{1+{{e}^{\dfrac{1}{x}}}}=0\]
As the domain of the above function is R-{0} therefore we can directly find the limit at x tends to zero, as function is same in both directions,
L.H.L. ( Left Hand Limit) \[=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x{{e}^{\dfrac{1}{x}}}}{1+{{e}^{\dfrac{1}{x}}}}\]
If we take the term \[{{e}^{\dfrac{1}{x}}}\] common from Numerator and denominator then we will get,
L.H.L. ( Left Hand Limit) \[=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x{{e}^{\dfrac{1}{x}}}}{{{e}^{\dfrac{1}{x}}}\left( 1+\dfrac{1}{{{e}^{\dfrac{1}{x}}}} \right)}\]
L.H.L. ( Left Hand Limit) \[=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{\left( 1+\dfrac{1}{{{e}^{\dfrac{1}{x}}}} \right)}\]
We will put the limits to get the final answer,
L.H.L. ( Left Hand Limit) \[=\dfrac{0}{\left( 1+\dfrac{1}{{{e}^{\dfrac{1}{0}}}} \right)}\]
As we all know that \[\dfrac{1}{0}\] is tending to\[\infty \], therefore above equation can be written as,
L.H.L. ( Left Hand Limit) \[=\dfrac{0}{\left( 1+\dfrac{1}{{{e}^{\infty }}} \right)}\]
L.H.L. ( Left Hand Limit) \[=\dfrac{0}{\left( 1+\dfrac{1}{\infty } \right)}\]
Also \[\dfrac{1}{\infty }\] is tending to 0 therefore we can write,
L.H.L. ( Left Hand Limit) \[=\dfrac{0}{\left( 1+0 \right)}\]
L.H.L. ( Left Hand Limit) \[=0\]
L.H.L. ( Left Hand Limit)= R.H.L. ( Right Hand Limit)
Therefore option (2) is correct.
Consider Option (3),
\[\underset{x\to 3}{\mathop{\lim }}\,{{(x-3)}^{\dfrac{1}{5}}} sgn (x-3)=0\]
As they have included the signum function in the problem, therefore we should know the definition of signum function given below,
Concept:
If, \[f(x)= sgn (x),\]then,
\[f(x)=-1\] \[For,x<0\]
\[f(x)=0\] \[For,x=0\]
\[f(x)=1\] \[For,x>0\]
Now to check whether the limit exist or not we will calculate the Left and Right hand limit accordingly,
L.H.L. ( Left Hand Limit )\[=\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,{{(x-3)}^{\dfrac{1}{5}}} sgn (x-3)\]
As we are finding the limit of negative side,
Put, \[x=3-h\]
As, \[x\to {{3}^{-}},h\to 0\]
L.H.L. ( Left Hand Limit )\[=\underset{h\to 0}{\mathop{\lim }}\,{{(3-h-3)}^{\dfrac{1}{5}}} sgn (3-h-3)\]
L.H.L. ( Left Hand Limit )\[=\underset{h\to 0}{\mathop{\lim }}\,{{(-h)}^{\dfrac{1}{5}}} sgn (-h)\]
L.H.L. ( Left Hand Limit )\[=\underset{h\to 0}{\mathop{\lim }}\,{{(-h)}^{\dfrac{1}{5}}}\left[ - sgn (h) \right]\]
As ‘h’ is tending to zero from the negative side of zero, therefore we can say that x<0 and from the definition of Signum function given above the value of signum function will be ‘-1’.
L.H.L. ( Left Hand Limit )\[=\underset{h\to 0}{\mathop{\lim }}\,{{(-h)}^{\dfrac{1}{5}}}\left[ -(-1) \right]\]
Now if we put the limits directly we will get,
L.H.L. ( Left Hand Limit )\[=\underset{h\to 0}{\mathop{\lim }}\,{{(-h)}^{\dfrac{1}{5}}}\times 1\]
L.H.L. ( Left Hand Limit )\[=0\]…………………………………. (i)
Also consider,
R.H.L. ( Right Hand Limit )\[=\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,{{(x-3)}^{\dfrac{1}{5}}} sgn (x-3)\]
As we are finding the limit of positive side,
Put, \[x=3+h\]
As, \[x\to {{3}^{+}},h\to 0\]
R.H.L. ( Right Hand Limit ) \[=\underset{h\to 0}{\mathop{\lim }}\,{{(3+h-3)}^{\dfrac{1}{5}}} sgn (3+h-3)\]
R.H.L. ( Right Hand Limit ) \[=\underset{h\to 0}{\mathop{\lim }}\,{{(h)}^{\dfrac{1}{5}}} sgn (h)\]
As ‘h’ is tending to zero from the positive side of zero, therefore we can say that x>0 and from the definition of Signum function given above the value of signum function will be ‘1’.
R.H.L. ( Right Hand Limit ) \[=\underset{h\to 0}{\mathop{\lim }}\,{{(h)}^{\dfrac{1}{5}}}(1)\]
Now if we put the limits directly we will get,
R.H.L. ( Right Hand Limit ) \[={{(0)}^{\dfrac{1}{5}}}(1)\]
R.H.L. ( Right Hand Limit ) \[=0\]…………………………………. (ii)
R.H.L. ( Right Hand Limit )= L.H.L. ( Left Hand Limit )
Therefore option (3) is correct.
Consider Option (4),
\[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}|x|}{x}=0\]
As they have included the modulus function in the problem therefore we should know the definition of modulus function given below,
Concept:
If f(x) =|x| then,
\[f(x)=-x\] \[at,x<0\]
\[f(x)=0\] \[at,x=0\]
\[f(x)=x\] \[at,x>0\]
By using the above concept we can write the left hand side of option 4 as shown below,
\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}\left( -x \right)}{x}\] \[at,x<0\]
\[L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}\left( x \right)}{x}\] \[at,x>0\]


Now consider,
L.H.L. ( Left Hand Limit )\[=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}\left( -x \right)}{x}\]
As we are finding the limit of negative side,
Put, \[x=0-h\]
As, \[x\to {{0}^{-}},h\to 0\]
L.H.L. ( Left Hand Limit )\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}\left[ -\left( 0-h \right) \right]}{0-h}\]
L.H.L. ( Left Hand Limit )\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}h}{-h}\]
If we put the limits directly we will get the indeterminate form given below,
L.H.L. ( Left Hand Limit )\[=\dfrac{0}{0}\]
Therefore to solve further we should use the L-Hospital’s rule given below,
L-Hospital’s Rule:
\[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]
By using the above rule we can write L.H.L. as,
L.H.L. ( Left Hand Limit )\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}{{\tan }^{-1}}h}{\dfrac{d}{dx}\left( -h \right)}\]
To solve further we should know the formula given below,
Formulae:

• 1. \[\dfrac{d}{dx}{{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}}\]

2. \[\dfrac{d}{dx}x=1\]

By using formulae given above we can write,
L.H.L. ( Left Hand Limit )\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{1+{{h}^{2}}}}{-1}\]
If we put the limits directly we will get,
L.H.L. ( Left Hand Limit )\[=\dfrac{\dfrac{1}{1+{{0}^{2}}}}{-1}\]
L.H.L. ( Left Hand Limit )\[=-1\]……………………………………….. (i)
Also consider,
R.H.L. ( Right Hand Limit )\[=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}\left( x \right)}{x}\]
As we are finding the limit of negative side,
Put, \[x=0+h\]
As, \[x\to {{0}^{+}},h\to 0\]
R.H.L. ( Right Hand Limit )\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}\left( 0+h \right)}{0+h}\]
R.H.L. ( Right Hand Limit )\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{-1}}h}{h}\]
If we put the limits directly we will get the indeterminate form given below,
R.H.L. ( Right Hand Limit )\[=\dfrac{0}{0}\]
Therefore to solve further we should use the L-Hospital’s rule given bellow,
L-Hospital’s Rule:
\[\underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}\]
By using above rule we can write L.H.L. as,
R.H.L. ( Right Hand Limit )\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}{{\tan }^{-1}}h}{\dfrac{d}{dx}h}\]
To solve further we should know the formula given below,
Formulae:
3. \[\dfrac{d}{dx}{{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}}\]
4. \[\dfrac{d}{dx}x=1\]
By using formulae given above we can write,
R.H.L. ( Right Hand Limit )\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{1+{{h}^{2}}}}{1}\]
If we put the limits directly we will get,
R.H.L. ( Right Hand Limit )\[=\dfrac{\dfrac{1}{1+{{0}^{2}}}}{1}\]
R.H.L. ( Right Hand Limit )\[=1\]……………………………………….. (ii)
From (i) and (ii) we can say that,

L.H.L. ( Left Hand Limit )\[\ne \] R.H.L. ( Right Hand Limit )
Which means limit does not exist at 0.
Therefore option (4) is incorrect.

Note: While solving this type of problem where you have to check the value of limits is correct or not, please try to define the functions whenever it is required otherwise your answer can be wrong.
For e.g.
If f(x) =|x| then,
\[f(x)=-x\] \[at,x<0\]
\[f(x)=0\] \[at,x=0\]
\[f(x)=x\] \[at,x>0\]