이정도 수학문제 다 푸실줄 아나요? +160 [10]

댓글(10)

• 윤성중학교대영유선애비

  
  

• 지원재경

  전 못 풉니다.

• 붕붕라이더

  일단 읽을 수 도 없음

• 숏커트

  분모, 분자 모두 무한대로 가니 간단히 0

• ?/zebra

  그건이미문제에있고 증명하라는거아닌가요?

• 120MS

  0ㄷㄷㄷ

• 돌샘희

  제일 쉬운 수학문제의 류가....임의수가 들어간 문제죠....

• 병신을보면짖는

  This image shows a math problem written in Korean. The problem asks to prove that for any integer n, the limit of frac{x^n}{e^x} as x approaches infinity is 0.
  
  Here's the problem and a solution:
  
  문제 (Problem)
  
  임의의 정수 n에 대하여 lim_{xtoinfty} frac{x^n}{e^x} = 0임을 보여라.
  
  (Prove that for any integer n, lim_{xtoinfty} frac{x^n}{e^x} = 0.)
  
  풀이 (Solution)
  
  To solve this, we can use L'H?pital's Rule. This rule is applicable when we have an indeterminate form like frac{infty}{infty} or frac{0}{0}. As x to infty, both x^n and e^x approach infinity, so we have an indeterminate form of type frac{infty}{infty}.
  
  Step 1: Differentiate the numerator and the denominator.
  
  frac{d}{dx}(x^n) = nx^{n-1}
  
  frac{d}{dx}(e^x) = e^x
  
  Applying L'H?pital's Rule once:
  
  lim_{xtoinfty} frac{x^n}{e^x} = lim_{xtoinfty} frac{nx^{n-1}}{e^x}
  
  Step 2: Repeat the process.
  
  We can see that if n is a positive integer, we still have an indeterminate form of type frac{infty}{infty}. We can apply L'H?pital's Rule repeatedly, n times.
  
  After the first application, we get: lim_{xtoinfty} frac{nx^{n-1}}{e^x}
  
  After the second application, we get: lim_{xtoinfty} frac{n(n-1)x^{n-2}}{e^x}
  
  ...
  
  After the n-th application, the numerator becomes a constant:
  
  lim_{xtoinfty} frac{n(n-1)(n-2)cdots(1)x^0}{e^x} = lim_{xtoinfty} frac{n!}{e^x}
  
  Step 3: Evaluate the final limit.
  
  As x to infty, the denominator e^x grows infinitely large, while the numerator n! is a constant.
  
  Therefore,
  
  lim_{xtoinfty} frac{n!}{e^x} = frac{n!}{infty} = 0
  
  This proves the statement for any positive integer n.
  
  Note on negative integers:
  
  If n is a negative integer, let n = -m where m is a positive integer.
  
  The limit becomes:
  
  lim_{xtoinfty} frac{x^{-m}}{e^x} = lim_{xtoinfty} frac{1}{x^m e^x}
  
  As x to infty, both x^m and e^x approach infinity. Their product, x^m e^x, also approaches infinity.
  
  Therefore,
  
  lim_{xtoinfty} frac{1}{x^m e^x} = frac{1}{infty} = 0
  
  Conclusion:
  
  In all cases (positive, negative, or zero for n), the limit is 0. This shows that the exponential function e^x grows faster than any polynomial function x^n.

• [∴]술퍼마리오

  GPT가... 이렇데요...
  

• 蘇部長☆