Question

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

Answer 1

Apply L-Hospital’s rule, it tells us that if we have an indeterminate form i.e $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ all we have to do is to differentiate the numerator and the denominator and then take the limits. L hospitals are applicable only if the value of f and g are 0, where f and g are defined as functions.
In this question we also have to apply the Maclaurin series theorem to substitute the value of $\sin x,\cos x$ and ${e^x}$ where Maclaurin series is a power series that is used to calculate an approximation of a function f(0) for input values close to zero, only if one knows the successive derivative of the function at zero.
Maclaurin series is a special case series of Taylor’s series, Maclaurin series can be written as
$\sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( 0 \right)\dfrac{{{x^n}}}{{n!}} = f\left( 0 \right) + f'\left( 0 \right)x} + \dfrac{{f''\left( 0 \right)}}{{2!}}{x^2} + ........ + \dfrac{{{f^{\left( k \right)}}\left( 0 \right)}}{{k!}}{x^k} + ....$

Complete step by step answer:
We have to find the value of a, b, c from the limit function ${\lim _{x \to 0}}\dfrac{{a + b\sin x - \cos x + c{e^x}}}{{{x^3}}}$
Hence we can write the limit as:
$l = {\lim _{x \to 0}}\dfrac{{a + b\sin x - \cos x + c{e^x}}}{{{x^3}}}$
Now using Maclaurin series, we know the value of $\sin x$, $\cos x$and ${e^x}$ at the limit $x \to 0$
$\sin x = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + .......$
$\cos x = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + .......$
${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + ..........$
Now by substituting these values in the limit function, we can write

$$l = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{a + b\sin x - \cos x + c{e^x}}}{{{x^3}}}} \right) \\\ = \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{a + b\left( {x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + .......} \right) - \left( {1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + .......} \right) + c\left( {1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + ..........} \right)}}{{{x^3}}}} \right] \\\$$

Now from the limit function find the sum of constant terms; hence we can write

$$a - 1 + c = 0 \\\ a + c = 1 - - - (i) \\\$$

Now find the sum of the coefficient of $x$
$b + c = 0 - - - (ii)$
And for the coefficient of ${x^2}$

$$\dfrac{1}{2} + \dfrac{c}{2} = 0 \\\ c = - 1 \\\$$

Hence substitute the value of $c = - 1$ in equations (i) and (ii), we get the values

$$a = 2 \\\ b = 1 \\\ c = - 1 \\\$$

Now let’s find the value of limit using L-Hospital’s rule, differentiate numerator and denominator

$${\lim _{x \to 0}}\dfrac{{a + b\sin x - \cos x + c{e^x}}}{{{x^3}}} = a - 1 + c = 0 \\\ {\lim _{x \to 0}}\dfrac{{b\cos x + \sin x + c{e^x}}}{{3{x^2}}} = b + c = 0 \\\ {\lim _{x \to 0}}\dfrac{{ - b\sin x + \cos x + c{e^x}}}{{6x}} = 1 + c = 0 \\\ {\lim _{x \to 0}}\dfrac{{ - b\cos x - \sin x + c{e^x}}}{6} = \dfrac{{ - b + c}}{6} \\\$$

Hence by substituting the value of $b = 1$ and $c = - 1$ we get the value of the limit
$\dfrac{{ - b + c}}{6} = \dfrac{{ - 1 + \left( { - 1} \right)}}{6} = \dfrac{{ - 2}}{6} = - \dfrac{1}{3}$ is the value of the limit.

Note: If the given function is in the indeterminate form of $\dfrac{0}{0}$ then we apply the L-Hospital’s rule where we differentiate the numerator and the denominator of the function until we get a non-zero solution.