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# Which one of the following statements is true?A. If $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right).g\left( x \right)$and $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right)$exist, then $\underset{x\to c}{\mathop{\lim }}\,g\left( x \right)$exists.B. If $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right).g\left( x \right)$exists, then $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right)$and $\underset{x\to c}{\mathop{\lim }}\,g\left( x \right)$exist.C. If $\underset{x\to c}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)$and $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right)$ exist, then $\underset{x\to c}{\mathop{\lim }}\,g\left( x \right)$exist.D. If$\underset{x\to c}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)$exists, then $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right)$and $\underset{x\to c}{\mathop{\lim }}\,g\left( x \right)$exist.

Last updated date: 20th Jun 2024
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Hint: Use counter examples to verify the given statements. Try to make examples related with the functions modulus, greatest integer, logarithm etc. to get accurate results.

Here, we have to find a true statement from the given options. So, we will discuss from option A to D with examples. As we need to use counter examples for the given statements.
Option A. is given as
If $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right).g\left( x \right)$and $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right)$exist, then $\underset{x\to c}{\mathop{\lim }}\,g\left( x \right)$exists.
Here one counter example of the given statement is given as
\begin{align} & f\left( x \right)=x \\ & g\left( x \right)=[x] \\ \end{align}
So, $f\left( x \right).g\left( x \right)=x[x]$.
And let c=0, and check for the existence of given functions.
As, $f\left( x \right)=x$ is existing at $x\to 0$ and will give$\underset{x\to 0}{\mathop{\lim }}\,x=0$.
So, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$ exists.
We have$f\left( x \right).g\left( x \right)=x[x]$.
Apply limit $x\to 0$to$x[x]$.
For LHL,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,x[x]$
Here [x] will give value ‘-1’as x is lying between (-1,0).
So, expression will become
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,x\left( -1 \right)=0$( As $x\to {{0}^{-}}$).
For RHL,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x[x]$
Here, [x] will give value ‘0’ as x is lying between (0,1).
Hence , we get $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x(0)=0(0)=0$
So, LHL=RHL for $f\left( x \right)g\left( x \right)$at $x\to 0$.
Hence, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right).g\left( x \right)$exists.
Now, we have $g\left( x \right)=[x]$.
For LHL, we get $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,[x]$
So, LHL=-1
For RHL, we get
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,[x]$.
So, RHL=0.
Hence, $\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)$ does not exist.
So, option A. is incorrect as here $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right).g\left( x \right)$exists and $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right)$exist but $\underset{x\to c}{\mathop{\lim }}\,g\left( x \right)$ doesn’t exist.
Option B. is given as
If $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right).g\left( x \right)$exists, then $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right)$and $\underset{x\to c}{\mathop{\lim }}\,g\left( x \right)$exist.
We can take some counter example as explained above where
\begin{align} & f\left( x \right)=x,g\left( x \right)=[x] \\ & f\left( x \right)g\left( x \right)=x[x] \\ \end{align}
So, option B. is incorrect as well.
Here, option D is given as If$\underset{x\to c}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)$exists, then $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right)$and $\underset{x\to c}{\mathop{\lim }}\,g\left( x \right)$exist.
We can take counter example for this option as
f\left( x \right)=\left\{ \begin{align} & +1,x\ge 0 \\ & -1,x<0 \\ \end{align} \right.
g\left( x \right)=\left\{ \begin{align} & -1,x\ge 0 \\ & +1,x<0 \\ \end{align} \right.
where $f\left( x \right)$and $g\left( x \right)$will not exist at $x\to 0$as LHL and RHL will have different values at $x\to {{0}^{-}}$and $x\to {{0}^{+}}$ respectively.
Now, $f\left( x \right)+g\left( x \right)$is given as
f\left( x \right)+g\left( x \right)=\left\{ \begin{align} & 0,x\ge 0 \\ & 0,x<0 \\ \end{align} \right.
or
$f\left( x \right)+g\left( x \right)$= 0
Which exists at $x\to 0$as $x\to {{0}^{-}}$and $x\to {{0}^{+}}$will give the same value 0.
Hence, option D. is also incorrect as$\underset{x\to c}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)$exists, but $\underset{x\to c}{\mathop{\lim }}\,f\left( x \right)$and $\underset{x\to c}{\mathop{\lim }}\,g\left( x \right)$is not existing in the above example.
Therefore option C. is the correct answer from the given statements.

Note: One cannot relate$f\left( x \right)$, $g\left( x \right)$and $f\left( x \right)+g\left( x \right)$or$f\left( x \right).g\left( x \right)$for any limit $x\to c$directly.
So, taking counterexamples to these kinds of questions is the only way to find the correct answer.
One can get some examples for corrections of option A, B, C, D but statements used here are general statements. So, try to use modulus, fractional, greatest integer functions etc. to verify these kinds of statements.