Processing math: 100%

Monday, 15 December 2014

Evaluating limits

Conventional methods:
1. Algebraic simplification – cancelling common factors, expanding, multiplying by conjugates
2. Rationalization
3. Trigonometry identities
4. Special limits

\lim_{x \to 0} \frac{\sin x}{x} = 1 \\ \\ \lim_{x \to 0} (1+\frac{1}{x})^x = e
5. L'Hôpital's rule

\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
Used only when we have indeterminate forms, i.e. \frac{0}{0}, \frac{\infty}{\infty}, 0 \times \infty, \infty - \infty, 0 \times 0, 1 \times \infty and \infty \times 0.

6. Sandwich's theorem

Miscellaneous examples:



Without L'H rule:



Key idea: The conjugate of trinomial a+b-c is a+b+c, as one can see [(a+b)-c][(a+b)+c] = (a+b)^2 - c^2.

Special methods:
1. Stripping

Example:
\lim_{x \to 0} \frac{\sin(\sin(3x))}{\ln(1+\sin(\sin(5x)))} \\ \\ x \to 0, 3x \to 0, \sin(\sin(5x)) \to 0
We can "strip off" the outer sin in the numerator and \ln (1+\_) in the denominator because the numerator is multiplicatively the same as \sin (3x), whereas the denominator is multiplicatively the same as \sin(\sin{5x}).

Strip again, we have \lim_{x \to 0} \frac{\sin(3x)}{\sin(\sin(5x))} \\ \\ x \to 0, 3x \to 0, \sin(5x) \to 0
Finally, strip off the sines in both the numerator and the denominator.

\lim_{x \to 0} \frac{3x}{\sin(5x)}
Now it is clear the limit is \frac{3}{5}.

2. Approximations
\sin x \sim x, \tan x \sim x, \arctan x \sim x, \arcsin x \sim x \\ e^x - 1 \sim x \\ \ln(1+x) \sim x (for \: small \: x) \\ 1-\cos x \sim \frac{x^2}{2}, 1-\cos\sqrt x \sim \frac{x}{2}
Example 1:
\lim_{x \to 0} \frac{x-\sin x}{x^2 (e^x-1)} \stackrel{e^x - 1 \sim x}{=} \lim_{x \to 0} \frac{x-\sin x}{x^3} \stackrel{L'H}{=} \lim_{x \to 0} \frac {\frac{x^2}{2}}{3x^2} = \frac{1}{6}
Example 2:
\lim_{x \to 0} \frac{e^x - \sin x - 1}{1-\sqrt {1-x^2}} \stackrel{\sqrt {1-x^2} \sim 1 - \frac{x^2}{2}}{=} \lim_{x \to 0} \frac{e^x - \sin x - 1}{\frac{x^2}{2}} \\ \stackrel{L'H}{=} \lim_{x \to 0} \frac {e^x - \cos x}{x} \stackrel{L'H}{=} \lim_{x \to 0} (e^x + \sin x)=1
3. Taylor series expansion

No comments:

Post a Comment