Consider the given limit \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+{{a}^{3}} \right)+8{{e}^{{}^{1}/{}_{x}}}}{1+\left( 1-{{b}^{3}} \right){{e}^{{}^{1}/{}_{x}}}}=2\] , then find the value of a and b.
A) a=1, b=2
B) $a=1,b=-{{3}^{\dfrac{1}{3}}}$
C) $a=2,b=3\dfrac{1}{3}$
D) none of these

Hint: Apply L’Hospital rule to the given expression, which states that if value of $\underset{x\to c}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}$ will be in indeterminate from of $\dfrac{0}{0}or\dfrac{\infty }{\infty }$ , then we can differentiate numerator and denominator individually and hence apply the limit $x\to c$ again i.e. we need to solve $\underset{x\to c}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}$

Complete step-by-step answer:
Given expression in the problem is \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+{{a}^{3}} \right)+8{{e}^{{}^{1}/{}_{x}}}}{1+\left( 1-{{b}^{3}} \right){{e}^{{}^{1}/{}_{x}}}}=2\ldots \ldots \left( i \right)\]
So, we need to determine the value of a and b with the help of the above equation.
Hence, put $x\to 0$ to the expression of L.H.S in the equation $\left( i \right)$ . So, we get \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+{{a}^{3}} \right)+8{{e}^{{}^{1}/{}_{x}}}}{1+\left( 1-{{b}^{3}} \right){{e}^{{}^{1}/{}_{x}}}}=\dfrac{\left( 1+{{a}^{3}} \right)+8{{e}^{{}^{1}/{}_{0}}}}{1+\left( 1-{{b}^{3}} \right){{e}^{{}^{1}/{}_{0}}}}\]
As we know $\dfrac{1}{0}\to \infty $ and ${{e}^{\infty }}\to \infty $ , as ${{e}^{x}}$ is an increasing function in $\left( -\infty ,\infty \right)$ .So, we can get expression of the limit as,
\[\dfrac{\left( 1+{{a}^{3}} \right)+8\left( \infty \right)}{1+\left( 1-{{b}^{3}} \right)\left( \infty \right)}=\dfrac{\infty }{\infty }\]
Hence, the given limit in the problem is of an indeterminate from i.e. $\dfrac{\infty }{\infty }$ . So, we need to simplify it before putting the limit to it. As indeterminate form of limit is $\dfrac{\infty }{\infty }$,
So we can use L’Hospital rule with it, which is given as:-
L’Hospital rule state that if value of $\underset{x\to c}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}$ is of the form $\dfrac{0}{0}or\dfrac{\infty }{\infty }$ , then, we can differentiate $f\left( x \right)$ and $g\left( x \right)$ individually and hence, try to put limit to the expression again i.e. we get expression as $\underset{x\to c}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}$ if $\underset{x\to c}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}$ is of the form $\dfrac{0}{0}or\dfrac{\infty }{\infty }$. And the same rule can be applied again, if we get an indeterminate form of limit $\left( \dfrac{0}{0}or\dfrac{\infty }{\infty } \right)$ again. So, applying the L’Hospital rule in the equation $\left( i \right)$ , we get
\[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( 1+{{a}^{3}} \right)+8{{e}^{{}^{1}/{}_{x}}}}{1+\left( 1-{{b}^{3}} \right){{e}^{{}^{1}/{}_{x}}}}=2\]
\[\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left[ \left( 1+{{a}^{3}} \right)+8{{e}^{{}^{1}/{}_{x}}} \right]}{\dfrac{d}{dx}\left[ 1+\left( 1-{{b}^{3}} \right){{e}^{{}^{1}/{}_{x}}} \right]}=2\]
\[\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( 1+{{a}^{3}} \right)+\dfrac{d}{dx}\left( 8{{e}^{{}^{1}/{}_{x}}} \right)}{\dfrac{d}{dx}\left( 1 \right)+\dfrac{d}{dx}\left( \left( 1-{{b}^{3}} \right){{e}^{{}^{1}/{}_{x}}} \right)}=2\]
\[\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left( 1+{{a}^{3}} \right)+8\dfrac{d}{dx}{{e}^{{}^{1}/{}_{x}}}}{\dfrac{d}{dx}\left( 1 \right)+\left( 1-{{b}^{3}} \right)\dfrac{d}{dx}{{e}^{{}^{1}/{}_{x}}}}=2\ldots \ldots \left( ii \right)\]
Now, as we know $\dfrac{d}{dx}\left(\text{constant}\right)=0$, $\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}$, $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$
So, we can use the above relation to simplify the equation (ii). And we need to use chain rule as well with the differentiation of ${{e}^{{}^{1}/{}_{x}}}$ , chain rule is given as,
$\left( f\left( g\left( x \right) \right)'=f'\left( g\left( x \right) \right)g'\left( x \right) \right)\ldots \ldots \left( iii \right)$
So, we apply chain rule with composite functions, as ${{e}^{{}^{1}/{}_{x}}}$ , is a composite relation because it is formed with the help of two functions:- exponential and algebraic both. So, we can get differentiate of ${{e}^{{}^{1}/{}_{x}}}$ with the help of equation $\left( iii \right)$ and using the identities given above i.e. $\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}$ and $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$. So, we get
$\begin{align}
  & \dfrac{d}{dx}\left( {{e}^{{}^{1}/{}_{x}}} \right)={{e}^{{}^{1}/{}_{x}}}.\dfrac{d}{dx}\left( \dfrac{1}{x} \right)={{e}^{{}^{1}/{}_{x}}}\dfrac{d}{dx}\left( {{x}^{-1}} \right) \\
 & \Rightarrow \dfrac{d}{dx}{{e}^{{}^{1}/{}_{x}}}={{-e}^{{}^{1}/{}_{x}}}.{{x}^{-2}}=\dfrac{{{-e}^{{}^{1}/{}_{x}}}}{{{x}^{2}}}\ldots \ldots \left( iv \right) \\
\end{align}$
Hence, we can get equation $\left( ii \right)$ as
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{-8{{e}^{{}^{1}/{}_{x}}}}{{{x}^{2}}}}{-\left( 1-{{b}^{3}} \right)\left( \dfrac{{{e}^{{}^{1}/{}_{x}}}}{{{x}^{2}}} \right)}=2$
Now, we can cancel out the term \[\dfrac{{{e}^{{}^{1}/{}_{x}}}}{{{x}^{2}}}\] from the numerator and denominator both, so, we get
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{-8}{-\left( 1-{{b}^{3}} \right)}=2$
As, there is no involvement of ‘x’ in the expression, so, we can remove $\underset{x\to 0}{\mathop{\lim }}\,$ , Hence we get $\begin{align}
  & \dfrac{-8}{-\left( 1-{{b}^{3}} \right)}=2 \\
 & \Rightarrow \dfrac{4}{1-{{b}^{3}}}=1 \\
\end{align}$
On cross multiplying the equation, we get $1-{{b}^{3}}=4$
\[1-4={{b}^{3}}\]
${{b}^{3}}=-3$
Taking cube root on both sides, we get
$b={{\left( -3 \right)}^{{}^{1}/{}_{3}}}$
Hence, the value of b is $-{{3}^{{}^{1}/{}_{3}}}$.
As the limit of the given expression is not depending at ‘a’, because we get the expression independent of ‘a’ after applying the L’Hospital Rule, it means, for any value of a, limit of the given expression will be 2, if $b=-{{3}^{{}^{1}/{}_{3}}}$.
Hence, option $\left( b \right)$ is the correct option.

Note: Another approach for solving this question would be that we can use expansion of ${{e}^{{}^{1}/{}_{x}}}$ as well. As expression ${{e}^{x}}$ is \[{{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\ldots \ldots \]
Replace x by $\dfrac{1}{x}$ , we get
\[{{e}^{{}^{1}/{}_{x}}}=1+\dfrac{1}{x}+\dfrac{1}{2!{{x}^{2}}}+\dfrac{1}{3!{{x}^{3}}}+\ldots \ldots \]
Now, put the expansion of ${{e}^{{}^{1}/{}_{x}}}$ to the given expression and hence find the value of ‘b’, and use the fact that limit of the given expression cannot tend to $\infty $ , it will be equal to 2.
Use the L’Hospital rule properly, as one may differentiate the whole expression, and not the denominator and numerator individually. So, be clear with the rule and apply it carefully as well. Don’t create confusion with the value of ‘a’ we can take any value of ‘a’, but the limit of the expression will not change, it will remain ‘2’ as we are not getting any term related to ‘a’ in the expression evaluated after applying L’Hospital rule.