MAT157Y1 Lecture : Special Limits.pdf

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5 Dec 2012
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Here"s a summary of the di erent limits that are important: f (0) = lim h 0 (or see the practice exam solutions). f (0+h) f (0) h exits and equals 2, as we calculated in tutorial, f f (0+h) f (0) +(0) = lim h 0+ tutorial, when i was thinking of still have to use the expression 4x + 1 when evaluating f (0), so: does not exist. lim x 0+ f (x)). As elnaz pointed out, you (i think i made a mistake in h. = lim h 0+ f (0+h) f (0) h h (2(0+h) 1) . You don"t actually need to compute this limit in the question. However, you could argue that f (0) does not exist using this calculation instead of using the continuity argument: lim x 0 , lim x 0+ f (x) = lim x 0 f (x) = lim x 0+

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