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# limits - x = √ x 1-1 x × √ x 1 1 √ x 1 1 = x 1-1 x...

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Definition : If f ( x ) becomes closer and closer to a single number L as x gets closer and closer to c from either side, then lim x c f ( x ) = L , which is read as “the limit of f ( x ) as x approaches c is L .” Examples: ( a ) lim x →- 1 3 x 2 + 5 = 3 × ( - 1) 2 + 5 = 8 . ( b ) lim x 2 p x 2 + 1 = p 2 2 + 1 = 5 ( c ) lim x 0 1 x 2 + 1 = 1 0 2 + 1 = 1 . (d) Suppose that f ( x ) = | x | x 6 = 0 1 x = 0 . Then lim x 0 f ( x ) = 0 . Note : lim x c f ( x ) relies on the values of f ( x ) at x near c , but may not have any connection to the value of f ( x ) at x = c . Replacement Theorem : If f ( x ) and g ( x ) are equal at all points except for at x = c . Then, lim x c f ( x ) = lim x c g ( x ) Example:
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Unformatted text preview: x = √ x + 1-1 x × √ x + 1 + 1 √ x + 1 + 1 = x + 1-1 x ( √ x + 1 + 1) = 1 √ x + 1 + 1 as long as x 6 = 0 (why?). Therefore, lim x → √ x + 1-1 x = lim x → 1 √ x + 1 + 1 = lim x → 1 lim x → ( √ x + 1 + 1) = 1 2 Extra Problems: Find the following limits ( a ) lim x → x 4 + 3 x 3-5 x 2 x , ( b ) lim x → 1 x 2-1 x-1 ( c ) lim x → 1 x-1 x 2-1 ( d ) lim x → 1 x-1 5 x-5...
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