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Can you prove that x → ∞ lim ( 3 4 2 x + 3 4 3 x ) x 1 = x → ∞ lim 3 4 3 × ( 1 + ( 3 4 3 3 4 2 ) x ) x 1
And why is ( 1 + a x ) x 1 for 0 < a < 1 is of the form 1 0 = 1 as x → ∞ ? The limit of a x as x → ∞ is 0 ?
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For the first question, we have that
( 3 4 2 x + 3 4 3 x ) x 1 = ( 3 4 3 x ( 1 + ( 3 4 3 3 4 2 ) x ) ) x 1 =
3 4 3 × ( 1 + ( 3 4 3 3 4 2 ) x ) x 1 ,
since ( 3 4 3 x ) x 1 = 3 4 3 . So if the first expression has a limit then it will be the same as that for the last expression.
For the second question, yes, since a x → 0 for 0 < a < 1 as x → ∞ we have ( 1 + a x ) x 1 being of the form 1 0 , which is not one of the indeterminate forms , so we can go ahead and conclude that the limit is 1 0 = 1 .
We can also look at it this way. If y = ( 1 + a x ) x 1 then ln ( y ) = x ln ( 1 + a x ) , which as x → ∞ goes to x ln ( 1 ) → 0 , and thus y → e 0 = 1 .
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This limit can be written as x → ∞ lim 3 4 3 × ( 1 + ( 3 4 3 3 4 2 ) x ) x 1 = 3 4 3 ,
since ( 1 + a x ) x 1 for 0 < a < 1 is of the form 1 0 = 1 as x → ∞ .
Comment: In general, for 0 < a ≤ b , x → ∞ lim ( a x + b x ) x 1 = b .