We're asked to find the limit (as n goes to infinity) of the following sequence: an=3n√4 −6 3n√2 + 92n√9 − 4 2n√3+4.
I thought that since the limit of the numerator exists, and since the limit of the denominator also exists and is non-zero, then the limit of the fraction should be 94.
But apperently, the limit of this sequence is 4 as n→+∞.
I don't understand why my approach is incorrect, nor why the limit of the sequence is 4.
Answer
It's just 1−6+91−4+4=4 because for example 413n→40=1.
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