Question

Which of the following is true ({.} denotes the fractional part of the function)?
(1).$\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{\log }_{e}}x}{\{x\}}=\infty $
(2).$\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\dfrac{x}{{{x}^{2}}-x-2}=\infty $
(3).$\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,\dfrac{x}{{{x}^{2}}-x-2}=-\infty $
(4).$\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{\log }_{0.5}}x}{\{x\}}=\infty $

Answer 1

Hint: To solve this problem we will solve all the options and check whether they are right or not.

Consider option (1),
$\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{\log }_{e}}x}{\{x\}}=\infty $
L.H.S. (Left Hand Side) = $\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{\log }_{e}}x}{\{x\}}$
Before solving it, we should know the limit of fractional part function which is given below,
Formula:
$\underset{x\to a}{\mathop{\lim }}\,\{x\}=a$ If the value of ‘a’ is a fraction which is between ‘0’ to ‘1’ always.
              = 0 if the value of ‘a’ is an integer as there is a fraction of ‘0’ in an integer.
Referring the above formula we can say that the value of {x} in L.H.S. will be in between 0 and 1,
Therefore after substituting the limits L.H.S. will become,

$L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\dfrac{{{\log }_{e}}\infty }{Any\text{ }value\text{ }between\text{ }0\text{ }and\text{ }1}$
As we all know that ${{\log }_{e}}\infty =\infty $,
$\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\dfrac{\infty }{Any\text{ }value\text{ }between\text{ }0\text{ }and\text{ }1}$
$\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\infty $
$\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=R.H.S.(Right\text{ }Hand\text{ }Side)$
$\therefore \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{\log }_{e}}x}{\{x\}}=\infty $
\[\therefore \] Option (1) is correct……………………………………………….. (i)
Consider Option (2)
$\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\dfrac{x}{{{x}^{2}}-x-2}=\infty $
$\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\dfrac{x}{{{x}^{2}}-x-2}$
By rearranging the denominator we will get,
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{h\to {{2}^{+}}}{\mathop{\lim }}\,\dfrac{x}{{{x}^{2}}-2x+x-2}\]
Taking x common in denominator we can write,
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{h\to {{2}^{+}}}{\mathop{\lim }}\,\dfrac{x}{x(x-2)+x-2}\]
Taking (x-2) common in denominator we can write,
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{h\to {{2}^{+}}}{\mathop{\lim }}\,\dfrac{x}{(x-2)(x+1)}\]
We have to find the limits in the positive side therefore assume,
X as $(2+h)$
As $x\to 2,h\to 0$
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{(2+h)}{(2+h-2)(2+h+1)}\]
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{(2+h)}{h\times (3+h)}\]
If we put the limits then we will get,
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\dfrac{(2+0)}{0\times (3+0)}\]
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\dfrac{2}{0}\]
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\infty \]
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=R.H.S.(Right\text{ }Hand\text{ }Side)\]
\[\therefore \underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\dfrac{x}{{{x}^{2}}-x-2}=\infty \]
\[\therefore \] Option (2) is correct……………………………………………….. (ii)
Consider Option (3),
$\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,\dfrac{x}{{{x}^{2}}-x-2}=-\infty $
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,\dfrac{x}{{{x}^{2}}-x-2}\]
By rearranging the denominator we will get,
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{h\to -{{1}^{-}}}{\mathop{\lim }}\,\dfrac{x}{{{x}^{2}}-2x+x-2}\]
Taking x common in denominator we can write,
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{h\to -{{1}^{-}}}{\mathop{\lim }}\,\dfrac{x}{x(x-2)+x-2}\]
Taking (x-2) common in denominator we can write,
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{h\to -{{1}^{-}}}{\mathop{\lim }}\,\dfrac{x}{(x-2)(x+1)}\]

We have to find the limits in the negative side therefore assume,
X as \[(-1-h)\]
As \[x\to -1,h\to 0\]
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{(-1-h)}{(-1-h-2)(-1-h+1)}\]
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{(-1-h)}{(-4-h)\times -h}\]
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-(1+h)}{-(4+h)\times -h}\]
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{h\to 0}{\mathop{\lim }}\,\left[ -\dfrac{(1+h)}{(4+h)\times h} \right]\]
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=-\underset{h\to 0}{\mathop{\lim }}\,\dfrac{(1+h)}{(4+h)\times h}\]
If we put the limits then we will get,
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=-\dfrac{(1+0)}{(4+0)\times 0}\]
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=-\dfrac{1}{0}\]
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=-\infty \]
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=R.H.S.(Right\text{ }Hand\text{ }Side)\]
\[\therefore \underset{x\to -{{1}^{-}}}{\mathop{\lim }}\,\dfrac{x}{{{x}^{2}}-x-2}=-\infty \]
\[\therefore \] Option (3) is correct……………………………………………….. (iii)

Consider option (4),
$\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{\log }_{0.5}}x}{\{x\}}=\infty $
\[L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{\log }_{0.5}}x}{\{x\}}\]
Before solving it, we should know the limit of fractional part function which is given below,
Formula:
$\underset{x\to a}{\mathop{\lim }}\,\{x\}=a$ If the value of ‘a’ is a fraction which is between ‘0’ to ‘1’ always.
              = 0 if the value of ‘a’ is an integer as there is a fraction of ‘0’ in an integer.
Referring the above formula we can say that the value of {x} in L.H.S. will be in between 0 and 1,
Therefore after substituting the limits L.H.S. will become,

\[L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=\dfrac{{{\log }_{0.5}}\infty }{Any\text{ }value\text{ }between\text{ }0\text{ }and\text{ }1}\]
To proceed further we should know the concept given below,
Concept:
\[{{\log }_{a}}x\] increases with the value of ‘x’ if ‘a>1’ and decreases with the value of ‘x’ if ‘a<1’
Now in the given example the value of ‘a’ is 0.5 which is less than 1.
Therefore as the value of ‘x’ increases the value of \[{{\log }_{a}}x\] decreases and after a certain value of x it starts becoming negative.
As ‘x’ is \[\infty \] therefore value of \[{{\log }_{0.5}}\infty \] will become \[-\infty \]from above concept,
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\text{ }=-\infty \]
\[\therefore L.H.S.\text{ }\left( Left\text{ }Hand\text{ }Side \right)\ne R.H.S.(Right\text{ }Hand\text{ }Side)\]
\[\therefore \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{\log }_{e}}x}{\{x\}}\ne \infty \]
\[\therefore \] Option (4) is not correct……………………………………………….. (iv)
From (i), (ii), (iii), and (iv) we can write our answer as given below,
 Option (1), (2), and (3) are correct.

Note: There are chances of committing mistakes in option (1) and (4) if you solve them without considering their bases as only the base concludes whether the answer is right or wrong.