Question

Find the value of $\lambda + n$, where $\lim_{x\to 0} \left( \frac{\cos^2 x - \cos x - e^x \cos x + e^x - \frac{x^3}{2}}{x^n} \right) = \lambda (\in \mathbb{R} - \{0\}); n \in \mathbb{N}.$

Answer 1

4.5

Answer 2

Let the given limit be $L$. $L = \lim_{x\to 0} \left( \frac{\cos^2 x - \cos x - e^x \cos x + e^x - \frac{x^3}{2}}{x^n} \right)$ We are given that $L = \lambda$, where $\lambda \in \mathbb{R} - \{0\}$, and $n \in \mathbb{N}$. For the limit to be a finite non-zero value, the numerator must have a Taylor series expansion around $x=0$ starting with a term proportional to $x^n$. The terms of order less than $n$ must cancel out.

Let the numerator be $N(x) = \cos^2 x - \cos x - e^x \cos x + e^x - \frac{x^3}{2}$. We use the Maclaurin series expansions for the functions involved: $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + O(x^8) = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + O(x^8)$ $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + O(x^7) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \frac{x^6}{720} + O(x^7)$

Expand each term in the numerator up to a sufficiently high order (let's aim for $x^6$ or $x^7$ since lower order terms are expected to cancel).

$\cos^2 x = \frac{1 + \cos(2x)}{2} = \frac{1}{2} + \frac{1}{2}\left(1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + O(x^8)\right)$ $= \frac{1}{2} + \frac{1}{2}\left(1 - 2x^2 + \frac{16x^4}{24} - \frac{64x^6}{720} + O(x^8)\right)$ $= \frac{1}{2} + \frac{1}{2} - x^2 + \frac{1}{3}x^4 - \frac{2}{45}x^6 + O(x^8) = 1 - x^2 + \frac{1}{3}x^4 - \frac{2}{45}x^6 + O(x^8)$

$e^x \cos x$: We can multiply the series expansions: $e^x \cos x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \frac{x^6}{720} + O(x^7)\right) \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + O(x^8)\right)$ Collecting terms up to $x^6$: Constant: $1 \times 1 = 1$ $x$: $x \times 1 = x$ $x^2$: $1 \times (-\frac{x^2}{2}) + \frac{x^2}{2} \times 1 = -\frac{x^2}{2} + \frac{x^2}{2} = 0$ $x^3$: $x \times (-\frac{x^2}{2}) + \frac{x^3}{6} \times 1 = -\frac{x^3}{2} + \frac{x^3}{6} = -\frac{2x^3}{6} = -\frac{x^3}{3}$ $x^4$: $1 \times \frac{x^4}{24} + \frac{x^2}{2} \times (-\frac{x^2}{2}) + \frac{x^4}{24} \times 1 = \frac{x^4}{24} - \frac{x^4}{4} + \frac{x^4}{24} = \left(\frac{1-6+1}{24}\right)x^4 = -\frac{4x^4}{24} = -\frac{x^4}{6}$ $x^5$: $x \times \frac{x^4}{24} + \frac{x^3}{6} \times (-\frac{x^2}{2}) + \frac{x^5}{120} \times 1 = \frac{x^5}{24} - \frac{x^5}{12} + \frac{x^5}{120} = \left(\frac{5-10+1}{120}\right)x^5 = -\frac{4x^5}{120} = -\frac{x^5}{30}$ $x^6$: $1 \times (-\frac{x^6}{720}) + \frac{x^2}{2} \times \frac{x^4}{24} + \frac{x^4}{24} \times (-\frac{x^2}{2}) + \frac{x^6}{720} \times 1 = -\frac{x^6}{720} + \frac{x^6}{48} - \frac{x^6}{48} + \frac{x^6}{720} = 0$ So, $e^x \cos x = 1 + x - \frac{x^3}{3} - \frac{x^4}{6} - \frac{x^5}{30} + O(x^7)$

Now substitute these expansions into $N(x)$: $N(x) = \left(1 - x^2 + \frac{1}{3}x^4 - \frac{2}{45}x^6 + O(x^8)\right)$ $- \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + O(x^8)\right)$ $- \left(1 + x - \frac{x^3}{3} - \frac{x^4}{6} - \frac{x^5}{30} + O(x^7)\right)$ $+ \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \frac{x^6}{720} + O(x^7)\right)$ $- \frac{x^3}{2}$

Collect coefficients for each power of $x$: Constant: $1 - 1 - 1 + 1 = 0$ $x$: $0 - 0 - 1 + 1 = 0$ $x^2$: $-1 - (-\frac{1}{2}) - 0 + \frac{1}{2} = -1 + \frac{1}{2} + \frac{1}{2} = 0$ $x^3$: $0 - 0 - (-\frac{1}{3}) + \frac{1}{6} - \frac{1}{2} = \frac{1}{3} + \frac{1}{6} - \frac{1}{2} = \frac{2+1-3}{6} = 0$ $x^4$: $\frac{1}{3} - \frac{1}{24} - (-\frac{1}{6}) + \frac{1}{24} = \frac{1}{3} - \frac{1}{24} + \frac{1}{6} + \frac{1}{24} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$ $x^5$: $0 - 0 - (-\frac{1}{30}) + \frac{1}{120} = \frac{1}{30} + \frac{1}{120} = \frac{4+1}{120} = \frac{5}{120} = \frac{1}{24}$ $x^6$: $-\frac{2}{45} - (-\frac{1}{720}) - 0 + \frac{1}{720} = -\frac{2}{45} + \frac{1}{720} + \frac{1}{720} = -\frac{32}{720} + \frac{1}{720} + \frac{1}{720} = \frac{-32+2}{720} = -\frac{30}{720} = -\frac{1}{24}$

So, $N(x) = \frac{1}{2}x^4 + \frac{1}{24}x^5 - \frac{1}{24}x^6 + O(x^7)$.

The limit is $L = \lim_{x\to 0} \frac{N(x)}{x^n} = \lim_{x\to 0} \frac{\frac{1}{2}x^4 + \frac{1}{24}x^5 - \frac{1}{24}x^6 + O(x^7)}{x^n}$. For the limit to be a finite non-zero value $\lambda$, the lowest power of $x$ in the numerator must match the power of $x$ in the denominator. The lowest power of $x$ in $N(x)$ is $x^4$. Thus, $n$ must be 4.

Substituting $n=4$: $L = \lim_{x\to 0} \frac{\frac{1}{2}x^4 + \frac{1}{24}x^5 - \frac{1}{24}x^6 + O(x^7)}{x^4} = \lim_{x\to 0} \left(\frac{1}{2} + \frac{1}{24}x - \frac{1}{24}x^2 + O(x^3)\right)$ As $x \to 0$, the limit is $\frac{1}{2}$. So, $\lambda = \frac{1}{2}$.

We are asked to find the value of $\lambda + n$. $\lambda + n = \frac{1}{2} + 4 = 4.5$.

Alternatively, we could use L'Hopital's rule, but the derivatives become increasingly complex. Let $N(x) = \cos^2 x - \cos x - e^x \cos x + e^x - \frac{x^3}{2}$. $N(0) = \cos^2 0 - \cos 0 - e^0 \cos 0 + e^0 - \frac{0^3}{2} = 1^2 - 1 - 1 \times 1 + 1 - 0 = 1 - 1 - 1 + 1 = 0$. $N'(x) = 2\cos x (-\sin x) - (-\sin x) - (e^x \cos x + e^x (-\sin x)) + e^x - \frac{3x^2}{2}$ $N'(x) = -2\sin x \cos x + \sin x - e^x \cos x + e^x \sin x + e^x - \frac{3x^2}{2}$ $N'(0) = -2(0)(1) + 0 - 1(1) + 1(0) + 1 - 0 = -1 + 1 = 0$. $N''(x) = -2(\cos^2 x - \sin^2 x) + \cos x - (e^x \cos x - e^x \sin x + e^x \sin x + e^x \cos x) + e^x - 3x$ $N''(x) = -2\cos(2x) + \cos x - (2e^x \cos x) + e^x - 3x$ $N''(0) = -2\cos(0) + \cos(0) - 2e^0 \cos(0) + e^0 - 0 = -2(1) + 1 - 2(1)(1) + 1 = -2 + 1 - 2 + 1 = -2$. Since $N''(0) \neq 0$, the lowest power of $x$ in the numerator expansion should be $x^2$. However, our Taylor series expansion showed the lowest power is $x^4$. Let's recheck the derivatives.

$N(x) = \cos^2 x - \cos x - e^x \cos x + e^x - \frac{x^3}{2}$ $N'(x) = 2\cos x(-\sin x) - (-\sin x) - (e^x \cos x - e^x \sin x) + e^x - \frac{3x^2}{2}$ $N'(x) = -\sin(2x) + \sin x - e^x \cos x + e^x \sin x + e^x - \frac{3x^2}{2}$ $N'(0) = 0 + 0 - 1(1) + 1(0) + 1 - 0 = -1 + 1 = 0$. Correct.

$N''(x) = -2\cos(2x) + \cos x - (e^x \cos x - e^x \sin x) + (e^x \sin x + e^x \cos x) + e^x - 3x$ $N''(x) = -2\cos(2x) + \cos x - e^x \cos x + e^x \sin x + e^x \sin x + e^x \cos x + e^x - 3x$ $N''(x) = -2\cos(2x) + \cos x + 2e^x \sin x + e^x - 3x$ $N''(0) = -2\cos(0) + \cos(0) + 2e^0 \sin(0) + e^0 - 0 = -2(1) + 1 + 2(1)(0) + 1 = -2 + 1 + 0 + 1 = 0$. Correct.

$N'''(x) = -2(-\sin(2x) \cdot 2) - \sin x + (2e^x \sin x + 2e^x \cos x) + e^x - 3$ $N'''(x) = 4\sin(2x) - \sin x + 2e^x \sin x + 2e^x \cos x + e^x - 3$ $N'''(0) = 4\sin(0) - \sin(0) + 2e^0 \sin(0) + 2e^0 \cos(0) + e^0 - 3 = 0 - 0 + 0 + 2(1)(1) + 1 - 3 = 2 + 1 - 3 = 0$. Correct.

$N^{(4)}(x) = 4(\cos(2x) \cdot 2) - \cos x + (2e^x \sin x + 2e^x \cos x) + (2e^x \cos x - 2e^x \sin x) + e^x$ $N^{(4)}(x) = 8\cos(2x) - \cos x + 4e^x \cos x + e^x$ $N^{(4)}(0) = 8\cos(0) - \cos(0) + 4e^0 \cos(0) + e^0 = 8(1) - 1 + 4(1)(1) + 1 = 8 - 1 + 4 + 1 = 12$.

Wait, according to the Taylor series $N(x) = \frac{N^{(4)}(0)}{4!} x^4 + O(x^5)$. The coefficient of $x^4$ is $\frac{12}{4!} = \frac{12}{24} = \frac{1}{2}$. This matches the Taylor series expansion result.

The limit is $L = \lim_{x\to 0} \frac{N(x)}{x^n}$. Since $N(x) = \frac{1}{2}x^4 + O(x^5)$, for the limit to be a non-zero finite value, we must have $n=4$. $L = \lim_{x\to 0} \frac{\frac{1}{2}x^4 + O(x^5)}{x^4} = \lim_{x\to 0} (\frac{1}{2} + O(x)) = \frac{1}{2}$. So, $n=4$ and $\lambda = \frac{1}{2}$.

The value of $\lambda + n = \frac{1}{2} + 4 = 4.5$.

The final answer is $\lambda + n = 4.5$.

The final answer is $\boxed{4.5}$.