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Need help in evaluating this limit
- Thread starter mathuser
- Start date
Have you heard of Stirling's formula?.
\(\displaystyle x!\sim x^{x}\cdot e^{-x}\sqrt{2\pi x}\)
\(\displaystyle \displaystyle\lim_{x\to \infty}\frac{ln(x^{x}\cdot e^{-x}\cdot\sqrt{2\pi x})}{xln(x)}\)
\(\displaystyle \displaystyle\lim_{x\to \infty}\frac{xln(x)-x+\frac{1}{2}ln(2\pi x)}{xln(x)}\)
\(\displaystyle \displaystyle\lim_{x\to \infty}\left[1-\frac{1}{ln(x)}+\frac{ln(2\pi x)}{2xln(x)}\right]\)
\(\displaystyle =\fbox{1}\)
\(\displaystyle x!\sim x^{x}\cdot e^{-x}\sqrt{2\pi x}\)
\(\displaystyle \displaystyle\lim_{x\to \infty}\frac{ln(x^{x}\cdot e^{-x}\cdot\sqrt{2\pi x})}{xln(x)}\)
\(\displaystyle \displaystyle\lim_{x\to \infty}\frac{xln(x)-x+\frac{1}{2}ln(2\pi x)}{xln(x)}\)
\(\displaystyle \displaystyle\lim_{x\to \infty}\left[1-\frac{1}{ln(x)}+\frac{ln(2\pi x)}{2xln(x)}\right]\)
\(\displaystyle =\fbox{1}\)
thanks for solving !!
Have you heard of Stirling's formula?.
\(\displaystyle x!\sim x^{x}\cdot e^{-x}\sqrt{2\pi x}\)
\(\displaystyle \displaystyle\lim_{x\to \infty}\frac{ln(x^{x}\cdot e^{-x}\cdot\sqrt{2\pi x})}{xln(x)}\)
\(\displaystyle \displaystyle\lim_{x\to \infty}\frac{xln(x)-x+\frac{1}{2}ln(2\pi x)}{xln(x)}\)
\(\displaystyle \displaystyle\lim_{x\to \infty}\left[1-\frac{1}{ln(x)}+\frac{ln(2\pi x)}{2xln(x)}\right]\)
\(\displaystyle =\fbox{1}\)
\(\displaystyle \displaystyle\lim_{x\to \infty}\frac{ln(x!)}{ln(x^{x})}\)
Rewrite ln(n!) as the sum of logs, and rewrite the denominator:
\(\displaystyle \displaystyle\lim_{x\to \infty}\frac{ln(x) \ + \ ln(x - 1) \ + \ ln(x - 2) \ + ... + \ ln(2) \ + ln(1)}{xln(x)}\)
Take the derivatives of the (sum of the) terms in the numerator and of the expression in the denominator:
\(\displaystyle \displaystyle\lim_{x\to \infty}\frac{\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2} + ...}{ln(x) + 1}\)
The new numerator is the sum of fractions with a descending order of
consecutive integers for its consecutive denominators.
\(\displaystyle \text{Let a and b belong to the set of Real numbers.}\)
If you can show, for appropriate a and b, that
\(\displaystyle ln(n) - a \ < \ new \ numerator \ < \ ln(n) + b,\ then \ the \ limit = 1 \)
\(\displaystyle will \ follow \ as \ n \to \infty.\)
?