Does #a_n=(8000n)/(.0001n^2) #converge? If so what is the limit?

1 Answer
Morgan
Dec 31, 2016

The sequence converges to 0.

Explanation:

We can evaluate the limit of this series by looking at dominant terms. The dominant term in the numerator is #n#, while the dominant term in the denominator is #n^2#.

#lim_(n->oo)(8000n)/(0.0001n^2)#

#=>lim_(n->oo)n/n^2#

#=>lim_(n->oo)1/n#

As #n# approaches infinity, #1/n# approaches #0#.

#=>lim_(n->oo)1/n=0#

#:.# The sequence converges to 0.

Alternatively, you can use L'Hospital's rule to simplify, though this takes a bit more work. It will yield the same result. By this rule, if:

#lim(f(x))/(g(x))=0/0# or #(+-oo)/(+-oo)#, then #lim(f(x))/(g(x))=lim(f'(x))/(g'(x))#

#lim_(n->oo)(8000n)/(0.0001n^2)=>(oo)/(oo)#

Take the derivative of the numerator and denominator:

#lim_(n->oo)=(8000)/(0.0002n)#

As #n# approaches infinity, #8000/(0.0002n)# approaches #0#. Essentially, you are dividing #8000# by bigger and bigger numbers until the number in the denominator is so large that the quotient is #~~0#.

Related questions
See all questions in Infinite Sequences
Impact of this question
1538 views around the world
You can reuse this answer
Creative Commons License