Coalesce - x →∞ ln x x − 1 = ” ∞ ∞ ” = lim x...

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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 θ → tan θ − θ θ 3 = ” ” = lim θ → sec 2 θ − 1 3 θ 2 = = lim θ → 2 sec 2 θ tan θ 6 θ = lim θ → sec 2 θ tan 2 θ + 2 sec 4 θ 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 π = 2 6 = 0 this is wrong : lim x → π − sin x 1 − cos x 6 = lim x → π − cos x sin x = −∞ 2...
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Coalesce - x →∞ ln x x − 1 = ” ∞ ∞ ” = lim x...

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