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: Recall two standard limits.
\[\lim_{x\to 0}\frac{\cos x - 1}{x^2} = -\frac{1}{2}, \qquad \lim_{x\to 0}\frac{e^x-1}{x}=1\]

Step 2: Rewrite the second factor using these.
\[\cos x - e^x = (\cos x - 1) - (e^x-1)\]
so
\[\lim_{x\to 0}\frac{\cos x - e^x}{x} = \lim_{x\to 0}\frac{\cos x-1}{x} - \lim_{x\to 0}\frac{e^x-1}{x} = 0 - 1 = -1\]

Step 3: Combine the two pieces.
\[\lim_{x\to 0}\frac{(\cos x-1)(\cos x - e^x)}{x^3} = \lim_{x\to 0}\frac{\cos x-1}{x^2}\cdot\lim_{x\to 0}\frac{\cos x-e^x}{x} = \left(-\frac{1}{2}\right)(-1) = \frac{1}{2}\]

Step 4: Conclusion.
Since this limit with \(n=3\) comes out finite and non-zero (any smaller n blows up to infinity, any larger n gives zero), the required value is
\[\boxed{n=3}\]