Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/200

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11. \lim_{x \to 0} x \cot \pi x. Ans. \frac{1}{\pi}. 18. \lim_{x \to 1} \left[ \frac{2}{x^2 - 1} - \frac{1}{x - 1} \right]. Ans. -1\frac{1}{2}
12. \lim_{y \to \infty} \frac{y}{e^{ay}}.   0. 19. \lim_{x \to 1} \left[ \frac{1}{\log x} - \frac{x}{\log x} \right].  
13. \lim_{x \to \frac{\pi}{2}} (\pi - 2x) \tan x.   2. 20. \lim_{\theta \to \frac{\pi}{2}} [\sec \theta - \tan \theta].   0.
14. \lim_{x \to \infty} x \sin \frac{a}{x}.   a. 21. \lim_{\phi \to 0} \left[ \frac{2}{\sin^2 \phi} - \frac{1}{1 - \cos \phi} \right].   \frac{1}{2}.
15. \lim_{x \to 0} x^n \log x. [n positive.]   0. 22. \lim_{y \to 1} \left[ \frac{y}{y - 1} - \frac{1}{\log y} \right].   \frac{1}{2}.
16. \lim_{\theta \to \frac{\pi}{4}} (1 - \tan \theta) \sec 2\theta.   1. 23. \lim_{z \to 0} \left[ \frac{\pi}{4z} - \frac{\pi}{2z(e^{\pi z} + 1)} \right].   \frac{\pi^2}{8}.
17. \lim_{\phi \to a} (a^2 - \phi^2) \tan \frac{\pi \phi}{2a}.   \frac{4 a^2}{\pi}.

115. Evaluation of the indeterminate forms 0^0, 1^{\infty}, \infty^0. Given a function of the form

  f(x)^{\phi(x)}.
In order that the function shall take on one of the above three forms, we must have for a certain value of x
  f(x) = 0, \phi(x) = 0, \text{giving} 0^0;
or, f(x) = 1, \phi(x) = \infty, \text{giving} 1^{\infty};
or, f(x) = \infty, \phi(x) = 0, \text{giving} {\infty}^0,
Let y = f(x)^{\phi(x)};
taking the logarithm of both sides,
  \log y = \phi(x) \log f(x).
In any of the above cases the logarithm of y (the function) will take on the indeterminate form
  0 \cdot \infty.

Evaluating this by the process illustrated in §113 gives the limit of the logarithm of the function. This being equal to the logarithm of the limit of the function, the limit of the function is known.[1]

Illustrative Example 1. Evaluate x^x when x = 0.

Solution. This function assumes the indeterminate form 0^0 for x = 0.
Let \ y =\ x^x;
then \ \log y = x \log x = 0 \cdot -\infty, when x = 0.
By § 113, \ \log y \frac{\log x}{\frac{1}{x}} = \frac{-\infty}{\infty}, when x = 0.
  1. Thus, if \lim \log_e y = a, then y = e^a.