Question

The integer n for which $\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cos x - 1} \right)\left( {\cos x - {e^x}} \right)}}{{{x^n}}}$ is a finite nonzero number is?
A. 3
B. 6
C. 9
D. 12

Answer 1

First we are required to find $\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cos x - 1} \right)\left( {\cos x - {e^x}} \right)}}{{{x^n}}}$. Note that, $\cos \theta = 1 - 2{\sin ^2}\dfrac{\theta }{2}$. Therefore, write $\left( {\cos x - 1} \right)$ as $- 2{\sin ^2}\left( {\dfrac{x}{2}} \right)$. Again, we know that $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin x}}{x}} \right) = 1$. Hence, substitute $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\cos x - 1}}{{{x^2}}}} \right) \times \left( {\dfrac{{\cos x - {e^x}}}{{{x^{n - 2}}}}} \right)$ with $- \dfrac{1}{2} \times \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^2} \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\cos x - {e^x}}}{{{x^{n - 2}}}}} \right)$ and proceed.

Formula used: $\cos \theta = 1 - 2{\sin ^2}\dfrac{\theta }{2}$
$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin x}}{x}} \right) = 1$

Complete step-by-step solution:
First we are required to find $\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cos x - 1} \right)\left( {\cos x - {e^x}} \right)}}{{{x^n}}}$.
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cos x - 1} \right)\left( {\cos x - {e^x}} \right)}}{{{x^n}}}$
$= \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\cos x - 1}}{{{x^2}}}} \right) \times \left( {\dfrac{{\cos x - {e^x}}}{{{x^{n - 2}}}}} \right)$
Since $\cos \theta = 1 - 2{\sin ^2}\dfrac{\theta }{2}$ ,
$= \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{ - 2{{\sin }^2}\left( {\dfrac{x}{2}} \right)}}{{4 \times {{\left( {\dfrac{x}{2}} \right)}^2}}}} \right) \times \left( {\dfrac{{\cos x - {e^x}}}{{{x^{n - 2}}}}} \right)$
$= - \dfrac{1}{2} \times \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^2} \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\cos x - {e^x}}}{{{x^{n - 2}}}}} \right)$ …….(1)
Again we know that, $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin x}}{x}} \right) = 1$,
Therefore, $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)$=1,
Substituting the value in (1), we get,
$= - \dfrac{1}{2} \times \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^2} \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\cos x - {e^x}}}{{{x^{n - 2}}}}} \right)$
$= - \dfrac{1}{2} \times 1 \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\cos x - {e^x}}}{{{x^{n - 2}}}}} \right)$
$= - \dfrac{1}{2} \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\cos x - {e^x}}}{{{x^{n - 2}}}}} \right)$
Now, $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\cos x - {e^x}}}{{{x^{n - 2}}}}} \right)$is in $\dfrac{0}{0}$ form. Therefore, we can use L’ Hospital Rule and differentiate both the numerator and denominator with respect to x.
So, we can write $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\cos x - {e^x}}}{{{x^{n - 2}}}}} \right)$ as $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{ - \sin x - {e^x}}}{{(n - 2){x^{n - 3}}}}} \right)$ with the help of L’ Hospital Rule.
Therefore, $- \dfrac{1}{2} \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\cos x - {e^x}}}{{{x^{n - 2}}}}} \right)$
$= - \dfrac{1}{2} \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{ - \sin x - {e^x}}}{{(n - 2){x^{n - 3}}}}} \right)$
$= - \dfrac{1}{{2(n - 2)}} \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{ - \sin x - {e^x}}}{{{x^{n - 3}}}}} \right)$
Substituting x=0 in the numerator, we get
$= - \dfrac{1}{{2(n - 2)}} \times - 1 \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{{x^{n - 3}}}}} \right)$
$= \dfrac{1}{{2(n - 2)}}\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{1}{{{x^{n - 3}}}}} \right)$
Now, according to the question, the limit will have finite non zero value if
${x^{n - 3}} = 1$
$\Rightarrow {x^{n - 3}} = {x^0}$
$\Rightarrow n - 3 = 0$
$\Rightarrow n = 3$
Hence, the integer n for which $\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cos x - 1} \right)\left( {\cos x - {e^x}} \right)}}{{{x^n}}}$ is a finite nonzero number, is 3.

Therefore, the correct answer is option (A).

Note: Note the few important formulae of limits:
$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin x}}{x}} \right) = 1$
$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\tan x}}{x}} \right) = 1$
$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^x} - 1}}{x}} \right) = 1$
$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{a^x} - 1}}{x}} \right) = {\log _e}a$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\log (1 + x)}}{x} = 1$
$\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\dfrac{1}{x}}} = e$
$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \dfrac{1}{x}} \right)^x} = e$
$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \dfrac{a}{x}} \right)^x} = {e^a}$