Thursday, 5 December 2019

real analysis - Limit of sequence is indeterminate.



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 limkk4(1149)k and limkk9(949)k



or in general limkkark where |r|<1 and a0.


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