Question

The values of $a,b$ and $c$ such that $\mathop {\lim }\limits_{x \to 0} \dfrac{{ax{e^x} - b\log \left( {1 + x} \right) + cx{e^{ - x}}}}{{{x^2}\sin x}} = 2$
A. $a = 3,b = 12,c = 9$
B. $a = 2,b = 4,c = 2$
C. $a = 2,b = 10,c = 8$
D, $a = 3,b = - 12,c = - 9$

Answer 1

This problem deals with solving the limit with L’Hospital’s rule. The L’Hospital’s rule is applied to a limit when the limit is in indeterminate form. This is done by differentiating the numerator and the denominator and then limit is applied again, which is given by:
$\Rightarrow \mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{{f'(a)}}{{g'(a)}}$
Here basic derivatives are also used such as:
$\Rightarrow \dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$
$\Rightarrow \dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$

Complete step-by-step answer:
Using the L’Hospital’s rule to the given limit.
Consider the given limit, as given below:
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{ax{e^x} - b\log \left( {1 + x} \right) + cx{e^{ - x}}}}{{{x^2}\sin x}} = 2$
We know that the limit of $\sin x$, when $x$ tends to zero is equal to $x$, which is given below:
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \sin x = x$
Also the limit of $\log \left( {1 + x} \right)$, when $x$ tends to zero is equal to $x$, which is given below:
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \log \left( {1 + x} \right) = x$
Now applying these limits on the above given limit, as given below:
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{ax{e^x} - b\log \left( {1 + x} \right) + cx{e^{ - x}}}}{{{x^2}\sin x}} = 2$
Substituting the considered and obtained limits, $\mathop {\lim }\limits_{x \to 0} \log \left( {1 + x} \right) = x$ and $\mathop {\lim }\limits_{x \to 0} \sin x = x$, as shown below:
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{ax{e^x} - b\left( x \right) + cx{e^{ - x}}}}{{{x^2}\left( x \right)}} = 2$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{ax{e^x} - bx + cx{e^{ - x}}}}{{{x^3}}} = 2$
Here taking the term $x$, in the numerator and denominator from the above limit, as given below:
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{x\left( {a{e^x} - b + c{e^{ - x}}} \right)}}{{{x^3}}} = 2$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{a{e^x} - b + c{e^{ - x}}}}{{{x^2}}} = 2$
Now if $x$ tends to zero, in the denominator the limit does not exist, hence the numerator should be equal to zero, which is given by:
$\Rightarrow \mathop {\lim }\limits_{x \to 0} a{e^x} - b + c{e^{ - x}} = 0$
We know that ${e^0} = 1$, hence the above limit is expressed as given below:
$\Rightarrow a\left( 1 \right) - b + c\left( 1 \right) = 0$
$\Rightarrow a - b + c = 0$
Now consider the limit $\mathop {\lim }\limits_{x \to 0} \dfrac{{a{e^x} - b + c{e^{ - x}}}}{{{x^2}}} = 2$, as this limit does not exist, hence applying the L’Hospital’s rule to the limit as shown below:
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{a{e^x} - 0 - c{e^{ - x}}}}{{2x}} = 2$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{a{e^x} - c{e^{ - x}}}}{{2x}} = 2$
Here even this limit does not exist as $x$ is present in the denominator, so the numerator should be equal to zero, which is given by:
$\Rightarrow \mathop {\lim }\limits_{x \to 0} a{e^x} - c{e^{ - x}} = 0$
$\Rightarrow a - c = 0$
Applying the L’Hospital’s rule to limit $\mathop {\lim }\limits_{x \to 0} \dfrac{{a{e^x} - c{e^{ - x}}}}{{2x}} = 2$, as given below:
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{a{e^x} + c{e^{ - x}}}}{2} = 2$
$\Rightarrow \dfrac{{a + c}}{2} = 2$
$\Rightarrow a + c = 4$
Now we have two equations and two variables, hence adding these two equations to get the values of $a$ and $c$, as given below:
$\Rightarrow a + c = 4$
$\Rightarrow a - c = 0$
$\Rightarrow 2a = 4$
$\therefore a = 2$
Substituting the value of $a = 2$, in the above equation, to get the value of $c$, as given below:
$\Rightarrow a + c = 4$
$\Rightarrow 2 + c = 4$
$\therefore c = 2$
From the first equation $a - b + c = 0$, finding the value of $b$, as given below:
$\Rightarrow b = a + c$
$\Rightarrow b = 2 + 2$
$\therefore b = 4$

Final Answer: The values of a, b and c are 2, 4 and 2 respectively.

Note:
Please note that in mathematics, more specifically in calculus, L’Hospital’s rule provides a technique to evaluate limits of indeterminate forms. Application of the rule often converts an indeterminate form to an expression that can be easily evaluated by substitution.
Also note that if $\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$ , then $\dfrac{d}{{dx}}\left( {{e^{ - x}}} \right) = - {e^{ - x}}$