Does \( a_{n} \) Diverge to \(\infty\)?

[Positronized]

Homework Statement

Given that $$b_{n}\rightarrow\infty$$ and $$\frac{a_{n}}{b_{n}}\rightarrow C$$ (where C>0) as $$n\rightarrow\infty$$, prove that $$a_{n}$$ must also diverge to $$\infty$$, that is, $$a_{n}\rightarrow\infty$$ as $$n\rightarrow\infty$$

Homework Equations

As above.

The Attempt at a Solution

I could easily deduce that $$\left\{a_{n}\right\}$$ cannot be bounded, otherwise the sequence of quotients will be null. Also, I deduced that $$\left\{a_{n}\right\}$$ cannot diverge to $$-\infty$$ as the limit of convergence cannot be positive in this case. However, I could not deduce that $$\left\{a_{n}\right\}$$ must be unbounded BOTH above and below, and so I cannot rule out the possibility of oscillation, yet.

Originally I said if $$\left\{a_{n}\right\}$$ oscillates AND also unbounded, then it is always possible to find a negative term no matter what the value of n is. Since $$\left\{b_{n}\right\}$$ is always positive for large n, then it is always possible to find the term of $$\left\{\frac{a_{n}}{b_{n}}\right\}$$ that is negative, therefore cannot converge to a positive limit. However it has become obvious that "unbounded" doesn't mean unbounded in both ways, i.e. sequences like {1,2,1,4,1,8,1,16,...} are unbounded and oscillating but will not give a negative term, so my argument isn't valid.

But apparently the easier approach is to use the epsilon definition, chosen the "right" epsilon and get the result directly without having to resort to proof by cases. I can't figure out what the "right" value is.

 

[Positronized]

Don't worry, I've got it (after thinking about it for a long time) by setting epsilon=l/2