Question

The integral value of \(n\) for which \[ \lim_{x\to 0}\frac{(\cos x-1)(\cos x-e^x)}{x^n} \] is a finite non-zero real number is

• A. \(4\)
• B. \(3\)
• C. \(2\)
• D. \(1\)

Hint:

For limits involving powers of \(x\), expand each function using Maclaurin series and identify the lowest non-zero degree term. The denominator must have the same power of \(x\) for the limit to be finite and non-zero.

Answer 1

The Correct Option is B

Step 1: Expand \(\cos x-1\) near \(x=0\).
Using the Maclaurin series, \[ \cos x=1-\frac{x^2}{2}+\frac{x^4}{24}+\cdots \] Therefore, \[ \cos x-1=-\frac{x^2}{2}+\frac{x^4}{24}+\cdots \] Hence, the lowest degree term is \[ \cos x-1\sim -\frac{x^2}{2} \]

Step 2: Expand \(\cos x-e^x\) near \(x=0\).
Using \[ e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots \] and \[ \cos x=1-\frac{x^2}{2}+\frac{x^4}{24}+\cdots \] we get \[ \cos x-e^x = \left(1-\frac{x^2}{2}+\cdots\right) - \left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots\right) \] \[ =-x-x^2-\frac{x^3}{6}+\cdots \] Hence, the lowest degree term is \[ \cos x-e^x\sim -x \]

Step 3: Find the lowest degree term of the numerator.
Multiplying the leading terms, \[ (\cos x-1)(\cos x-e^x) \sim \left(-\frac{x^2}{2}\right)(-x) \] \[ =\frac{x^3}{2} \] Thus, \[ (\cos x-1)(\cos x-e^x) = \frac{x^3}{2} +\text{higher order terms} \]

Step 4: Determine the value of \(n\).
The limit becomes \[ \lim_{x\to 0} \frac{\frac{x^3}{2}+\cdots}{x^n} \] For the limit to be finite and non-zero, the power of \(x\) in the denominator must match the lowest power of \(x\) in the numerator.
Therefore, \[ n=3 \]

Step 5: Verify the limit.
Substituting \(n=3\), \[ \lim_{x\to 0} \frac{(\cos x-1)(\cos x-e^x)}{x^3} = \frac{1}{2} \] which is finite and non-zero.
Hence, the required integral value of \(n\) is \[ 3 \]

Step 6: Final conclusion.
Therefore, \[ \boxed{3} \]