Question

If $\mathop {\lim }\limits_{x \to 0} \dfrac{{a + b\sin x - \cos x + c{e^x}}}{{{x^3}}}$ exists, find the value of a, b, c. Also find the limit.

Answer 1

Hint:
We can expand the trigonometric functions and exponential functions as its power series expansion. Then we can equate the coefficients of negative powers of x to zero. Then we can solve for a, b and c. Then we can substitute these values in the expression. Then we can apply L’ Hospital’s rule to find the limit.

Complete step by step solution:
We are given that $\mathop {\lim }\limits_{x \to 0} \dfrac{{a + b\sin x - \cos x + c{e^x}}}{{{x^3}}}$ exists.
Let $L = \mathop {\lim }\limits_{x \to 0} \dfrac{{a + b\sin x - \cos x + c{e^x}}}{{{x^3}}}$
We know that trigonometric functions and exponential functions can be written as its power series expansion. We know that the expansions are given as
 $\sin x = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - ...$
 $\cos x = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - ...$ and
 ${e^x} = 1 + \dfrac{x}{1} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + ...$
On substituting these expansions, we get the expression as
 $ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{a + b\left( {x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - ...} \right) - \left( {1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - ...} \right)x + c\left( {1 + \dfrac{x}{1} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + ...} \right)}}{{{x^3}}}$
As the limit exists, we can say that the coefficient of powers of x less than the power of denominator are zero.
On equating the constant terms, we get
 $ \Rightarrow a - 1 + c = 0$
On rearranging, we get
 $ \Rightarrow a + c = 1$ … (1)
Now we can equate the coefficients of x.
 $ \Rightarrow b + c = 0$ … (2)
Now we can equate the coefficients of ${x^2}$
 $ \Rightarrow 1 + c = 0$
On rearranging, we get
 $ \Rightarrow c = - 1$ … (3).
Now we have 3 equations and 3 unknowns
On substituting equation (3) in (2), we get,
 $ \Rightarrow b - 1 = 0$
On adding 1 on both sides we get
 $ \Rightarrow b = 1$
On substituting (3) in (1), we get
 $ \Rightarrow a - 1 = 1$
On adding 1 on both sides we get
 $ \Rightarrow a = 2$
Now we can substitute the values of a, b and c in the given expression.
 $ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{2 + \sin x - \cos x - {e^x}}}{{{x^3}}}$
Now we can directly apply the limit.
 $ \Rightarrow L = \dfrac{{2 + \sin 0 - \cos 0 - {e^0}}}{{{0^3}}}$
We know that ${e^0} = 1$, $\sin 0 = 0$ and $\cos 0 = 1$. So, we get,
 $ \Rightarrow L = \dfrac{{2 + 0 - 1 - 1}}{0}$
On simplification we get
 $ \Rightarrow L = \dfrac{0}{0}$
As the limit is $\dfrac{0}{0}$, we can apply L’ hospital’s rule.
We know that by L’ hospital’s rule, If $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{0}{0}$, then $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$
Here, we have $f\left( x \right) = 2 + \sin x - \cos x - {e^x}$ and $g\left( x \right) = {x^3}$
On taking the derivatives of both the functions, we get
 $f'\left( x \right) = \cos x + \sin x - {e^x}$ and $g'\left( x \right) = 3{x^2}$
So, the limit will become,
 $ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$
On substituting the values, we get
 \[ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos x + \sin x - {e^x}}}{{3{x^2}}}\]
On applying the limits, we get
 \[ \Rightarrow L = \dfrac{{\cos 0 + \sin 0 - {e^0}}}{{3 \times {0^2}}}\]
We know that ${e^0} = 1$, $\sin 0 = 0$ and $\cos 0 = 1$. So, we get
 $ \Rightarrow L = \dfrac{{1 + 0 - 1}}{0}$
On simplification we get
 $ \Rightarrow L = \dfrac{0}{0}$
As the limit is $\dfrac{0}{0}$, we can again apply L’ hospital’s rule.
Here, we have $f\left( x \right) = \cos x + \sin x - {e^x}$ and $g\left( x \right) = 3{x^2}$
On taking the derivatives of both the functions, we get
 $f'\left( x \right) = - \sin x + \cos x - {e^x}$ and $g'\left( x \right) = 6x$
So, the limit will become,
 $ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$
On substituting the values, we get
 \[ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - \sin x + \cos x - {e^x}}}{{6x}}\]
On applying the limits, we get
 \[ \Rightarrow L = \dfrac{{ - \sin 0 + \cos 0 - {e^0}}}{{6 \times {0^2}}}\]
We know that ${e^0} = 1$, $\sin 0 = 0$ and $\cos 0 = 1$. So, we get
 $ \Rightarrow L = \dfrac{{0 + 1 - 1}}{0}$
On simplification we get
 $ \Rightarrow L = \dfrac{0}{0}$
As the limit is $\dfrac{0}{0}$, we can again apply L’ hospital’s rule.
Here, we have $f\left( x \right) = - \sin x + \cos x - {e^x}$ and $g\left( x \right) = 6x$
On taking the derivatives of both the functions, we get
 $f'\left( x \right) = - \cos x - \sin x - {e^x}$ and $g'\left( x \right) = 6$
So, the limit will become,
 $ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$
On substituting the values, we get
 \[ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - \cos x - \sin x - {e^x}}}{6}\]
On applying the limits, we get
 \[ \Rightarrow L = \dfrac{{ - \cos 0 - \sin 0 - {e^0}}}{6}\]
We know that ${e^0} = 1$, $\sin 0 = 0$ and $\cos 0 = 1$. So, we get
 $ \Rightarrow L = \dfrac{{ - 1 - 0 - 1}}{6}$
On simplification, we get
 $ \Rightarrow L = \dfrac{{ - 2}}{6}$
Hence, we have
 $ \Rightarrow L = \dfrac{{ - 1}}{3}$

Therefore, the required value of limit is $\dfrac{{ - 1}}{3}$ and the values of a, b and c are 2, 1 and -1 respectively.

Note:
We can use only the L’ hospital’s rule when the limit is of the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$. We must identify the functions of numerator and denominator and find their derivative separately. We cannot take the \[{2^{nd}}\] derivative directly. We must check the limit after taking the \[{1^{st}}\] derivative before taking the \[{2^{nd}}\] derivative. We must take care that the coefficients of the same powers of the variable are added. As the denominators are the same for the same powers, we need to take the sum of the numerator and equate to zero.