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Question: Find the value of $\lambda + n$, where $$\lim_{x\to 0} \left( \frac{\cos^2 x - \cos x - e^x \cos x +...

Find the value of λ+n\lambda + n, where lim⁡x→0(cos⁡2x−cos⁡x−excos⁡x+ex−x32xn)=λ(∈R−{0});n∈N.\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

4.5

Explanation

Solution

The problem asks for the value of λ+n\lambda + n, where n∈Nn \in \mathbb{N} and λ∈R−{0}\lambda \in \mathbb{R}-\{0\} are defined by the limit: lim⁡x→0(cos⁡2x−cos⁡x−excos⁡x+ex−x32xn)=λ\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

Let the numerator be f(x)=cos⁡2x−cos⁡x−excos⁡x+ex−x32f(x) = \cos^2 x - \cos x - e^x \cos x + e^x - \frac{x^3}{2}. As x→0x \to 0, f(x)→cos⁡20−cos⁡0−e0cos⁡0+e0−032=12−1−1⋅1+1−0=1−1−1+1=0f(x) \to \cos^2 0 - \cos 0 - e^0 \cos 0 + e^0 - \frac{0^3}{2} = 1^2 - 1 - 1 \cdot 1 + 1 - 0 = 1 - 1 - 1 + 1 = 0.

Since the limit exists and is a non-zero real number λ\lambda, the limit must be of the form 00\frac{0}{0}, and the lowest power of xx in the Taylor series expansion of the numerator around x=0x=0 must be xnx^n. The coefficient of xnx^n will be λ\lambda.

We use the Taylor series expansions around x=0x=0: cos⁡x=1−x22!+x44!−x66!+x88!−⋯=1−x22+x424−x6720+x840320−…\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \dots = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \frac{x^8}{40320} - \dots ex=1+x+x22!+x33!+x44!+x55!+x66!+x77!+x88!+⋯=1+x+x22+x36+x424+x5120+x6720+x75040+x840320+…e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + \frac{x^7}{7!} + \frac{x^8}{8!} + \dots = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \frac{x^6}{720} + \frac{x^7}{5040} + \frac{x^8}{40320} + \dots

Expand each term in the numerator:

  1. cos⁡2x=(cos⁡x)2=(1−x22+x424−x6720+… )2\cos^2 x = (\cos x)^2 = \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \dots\right)^2 =12+(−x22)2+(x424)2+2(1)(−x22)+2(1)(x424)+2(1)(−x6720)+2(−x22)(x424)+2(−x22)(−x6720)+2(x424)(−x22)+…= 1^2 + \left(-\frac{x^2}{2}\right)^2 + \left(\frac{x^4}{24}\right)^2 + 2(1)\left(-\frac{x^2}{2}\right) + 2(1)\left(\frac{x^4}{24}\right) + 2(1)\left(-\frac{x^6}{720}\right) + 2\left(-\frac{x^2}{2}\right)\left(\frac{x^4}{24}\right) + 2\left(-\frac{x^2}{2}\right)\left(-\frac{x^6}{720}\right) + 2\left(\frac{x^4}{24}\right)\left(-\frac{x^2}{2}\right) + \dots =1−x2+x44+x412−x6360−x624+⋯=1−x2+3+112x4−2+15720x6+⋯=1−x2+13x4−17720x6+…= 1 - x^2 + \frac{x^4}{4} + \frac{x^4}{12} - \frac{x^6}{360} - \frac{x^6}{24} + \dots = 1 - x^2 + \frac{3+1}{12}x^4 - \frac{2+15}{720}x^6 + \dots = 1 - x^2 + \frac{1}{3}x^4 - \frac{17}{720}x^6 + \dots

  2. cos⁡x=1−x22+x424−x6720+x840320−…\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \frac{x^8}{40320} - \dots

  3. excos⁡xe^x \cos x: excos⁡x=1+x−x33−x46−x530+x7630+x82520+…e^x \cos x = 1 + x - \frac{x^3}{3} - \frac{x^4}{6} - \frac{x^5}{30} + \frac{x^7}{630} + \frac{x^8}{2520} + \dots

  4. ex=1+x+x22+x36+x424+x5120+x6720+x75040+x840320+…e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \frac{x^6}{720} + \frac{x^7}{5040} + \frac{x^8}{40320} + \dots

  5. −x32-\frac{x^3}{2}

Now, assemble f(x)=cos⁡2x−cos⁡x−excos⁡x+ex−x32f(x) = \cos^2 x - \cos x - e^x \cos x + e^x - \frac{x^3}{2}: f(x)=(1−x2+13x4−17720x6+6940320x8+… )f(x) = \left(1 - x^2 + \frac{1}{3}x^4 - \frac{17}{720}x^6 + \frac{69}{40320}x^8 + \dots\right) −(1−x22+x424−x6720+x840320−… )- \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \frac{x^8}{40320} - \dots\right) −(1+x−x33−x46−x530+x7630+x82520+… )- \left(1 + x - \frac{x^3}{3} - \frac{x^4}{6} - \frac{x^5}{30} + \frac{x^7}{630} + \frac{x^8}{2520} + \dots\right) +(1+x+x22+x36+x424+x5120+x6720+x75040+x840320+… )+ \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \frac{x^6}{720} + \frac{x^7}{5040} + \frac{x^8}{40320} + \dots\right) −x32- \frac{x^3}{2}

Collect coefficients for each power of xx: Constant: 1−1−1+1=01 - 1 - 1 + 1 = 0 x1x^1: 0−0−1+1=00 - 0 - 1 + 1 = 0 x2x^2: −1−(−12)−0+12=−1+12+12=0-1 - (-\frac{1}{2}) - 0 + \frac{1}{2} = -1 + \frac{1}{2} + \frac{1}{2} = 0 x3x^3: 0−0−(−13)+16−12=13+16−12=2+1−36=00 - 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 x4x^4: 13−124−(−16)+124=13−124+16+124=13+16=2+16=36=12\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} x5x^5: 0−0−(−130)+1120=130+1120=4+1120=5120=1240 - 0 - (-\frac{1}{30}) + \frac{1}{120} = \frac{1}{30} + \frac{1}{120} = \frac{4+1}{120} = \frac{5}{120} = \frac{1}{24} x6x^6: −17720−(−1720)−0+1720=−17720+1720+1720=−17+1+1720=−15720=−148-\frac{17}{720} - (-\frac{1}{720}) - 0 + \frac{1}{720} = -\frac{17}{720} + \frac{1}{720} + \frac{1}{720} = \frac{-17+1+1}{720} = -\frac{15}{720} = -\frac{1}{48}

So, f(x)=12x4+124x5−148x6−1720x7+5340320x8+…f(x) = \frac{1}{2}x^4 + \frac{1}{24}x^5 - \frac{1}{48}x^6 - \frac{1}{720}x^7 + \frac{53}{40320}x^8 + \dots

The lowest power of xx with a non-zero coefficient is x4x^4. Thus, n=4n=4. The coefficient of x4x^4 is 12\frac{1}{2}. So, λ=12\lambda = \frac{1}{2}.

The limit is lim⁡x→0f(x)xn=lim⁡x→012x4+124x5−148x6−…x4=lim⁡x→0(12+124x−148x2−… )=12\lim_{x\to 0} \frac{f(x)}{x^n} = \lim_{x\to 0} \frac{\frac{1}{2}x^4 + \frac{1}{24}x^5 - \frac{1}{48}x^6 - \dots}{x^4} = \lim_{x\to 0} \left(\frac{1}{2} + \frac{1}{24}x - \frac{1}{48}x^2 - \dots\right) = \frac{1}{2}. So, λ=12\lambda = \frac{1}{2} and n=4n=4.

The question asks for the value of λ+n\lambda + n. λ+n=12+4=4.5\lambda + n = \frac{1}{2} + 4 = 4.5.

The final answer is 12+4=1+82=92\frac{1}{2} + 4 = \frac{1+8}{2} = \frac{9}{2}.