Does #a_n=(8000n)/(.0001n^2) #converge? If so what is the limit?
1 Answer
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
#lim_(n->oo)(8000n)/(0.0001n^2)#
#=>lim_(n->oo)n/n^2#
#=>lim_(n->oo)1/n#
As
#=>lim_(n->oo)1/n=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