Question

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

Answer 1

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`.