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kiararamos · Oct 12, 2009, 09:59 AM

lim as x approaches pi (π)

Sin(π-x)
(π-x)

ebaines · Oct 12, 2009, 10:17 AM

Two ways to solve this:

1. If you substitute $u = \pi - x$ , you get:

$<br /> \lim _ {x \to \pi} \frac {sin(\pi - x)} {\pi - x} = \lim _{ u \to 0} \frac {sin(u)} u<br />$

This second form may look familiar to you.

2. Have you studied l'Hospita'ls rule yet? It's quite handy: the limit as x appraches c for a function composed of a numerator and a denominator is equal to the limit as x approaches c of the derivative of the numerator over the derivative of the denominator:

$<br /> \lim _{x \to c} \ \frac { f(x)} {g(x)} = \lim _{x \to c} \ \frac {f'(x)} {g'(x)}<br />$