We have to find the limit of the sequence ak=(k411k+k99k72k+1).
Here is my attempt:
(k411k+k99k72k+1)=k411k+k99k49k+1=(1149)kk5+(949)k1k9+1k949k
But as k→∞, both the numerator and the denominator tend to zero. So I am left with an indeterminate form. How can I avoid this?
Answer
Guide:
|k411k+k99k72k+1|≤k411k+k99k49k=k4(1149)k+k9(949)k
You can focus on finding limk→∞k4(1149)k and limk→∞k9(949)k
or in general limk→∞kark where |r|<1 and a≥0.
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