Coalesce - x ln x x 1 = = lim x 1 /x 1 = 0 lim x x 1 x 2 =...

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In the trig section we needed lim h 0 sin h h =” 0 0 ”=1 lim h 0 cos h 1 h =” 0 0 ”=1 We used a geometrical argument to solve it, but there are a lot of those functions with an indeterminant form 0 0 ”or” ” like lim x 1 ln x x 1 =” 0 0 lim x →∞ ln x x 1 =” here we can use L’Hospitals Rule : If f,g are diF. and g 0 ( x ) 6 = 0 near a (except possibly at a) and if lim x a f ( x ) = 0 and lim x a g ( x ) = 0 (i.e. we have ” 0 0 ”) or lim x a f ( x )= ±∞ and lim x a g ( x )= ±∞ (i.e. we have ” ”) then lim x a f ( x ) g ( x ) = lim x a f 0 ( x ) g 0 ( x ) I± the limit on the right side exists (or is ±∞ ) Remarks: L’Hospital says that the limit of the quotient a two functions is equal to the limit of the quotient of their derivatives , provided the given conditions It can also be used for one sided limits or if a = ±∞ – Proof for the case f ( a )=0= g ( a ): lim x a f 0 ( x ) g 0 ( x ) = f 0 ( a ) g 0 ( a ) = lim x a f ( x ) f ( a ) x a lim x a g ( x ) g ( a ) x a = lim x a f ( x ) f ( a ) g ( x ) g ( a ) = lim x a f ( x ) g ( x ) So now we can solve lim h 0 sin h h =” 0 0 ” = lim h 0 cos h 1 =1 lim h 0 cos h 1 h =” 0 0 ” = lim h 0 sin h 1 =0 lim x 1 ln x x 1 =” 0 0 ” = lim x 1 1 /x 1 =1 1
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Unformatted text preview: x ln x x 1 = = lim x 1 /x 1 = 0 lim x x 1 x 2 = = lim x 1 2 x = 0 lim x x 2 1 2 x 2 + 1 = = lim x 2 x 4 x = 1 / 2 lim x e x x 2 = = lim x e x 2 x = = lim x e x 2 = lim tan 3 = = lim sec 2 1 3 2 = = lim 2 sec 2 tan 6 = lim sec 2 tan 2 + 2 sec 4 6 = 1 / 3 lim x + x ln x = lim x + ln x 1 /x = = lim x + 1 /x 1 /x 2 = lim x + x = 0 But be careful! Since 1 cos = 2 6 = 0 this is wrong : lim x sin x 1 cos x 6 = lim x cos x sin x = 2...
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Coalesce - x ln x x 1 = = lim x 1 /x 1 = 0 lim x x 1 x 2 =...

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