# Thread: Sequences-Original Diverges, Inverse Converges?

1. ## Sequences-Original Diverges, Inverse Converges?

Suppose f satisfies $\displaystyle \lim_{x->0^+}f(x)=+\infty$. Is it possible the sequence {$\displaystyle f(\frac{1}{n})$} converges? Explain.
The answer to this question is "yes" (I don't know why), because the second part asks:
Find a function f that
supports this theory.
I don't really understand why the answer is yes. The only function I can think of that goes to infinty as x goes to 0 is $\displaystyle \frac{1}{x^n}$.

2. Originally Posted by WhoCares357
The answer to this question is "yes" (I don't know why), because the second part asks:

I don't really understand why the answer is yes. The only function I can think of that goes to infinty as x goes to 0 is $\displaystyle \frac{1}{x^n}$.

Great. So you found a function, $\displaystyle f(x)=\frac{1}{x^n}$ s.t $\displaystyle \lim_{x\rightarrow 0^+}f(x)=\infty\,\,and\,\,f\left(\frac{1}{n}\right )=n^n$ converges (and to zero)...so what's the problem, again?

Tonio

3. Originally Posted by tonio
Great. So you found a function, $\displaystyle f(x)=\frac{1}{x^n}$ s.t $\displaystyle \lim_{x\rightarrow 0^+}f(x)=\infty\,\,and\,\,f\left(\frac{1}{n}\right )=n^n$ converges (and to zero)...so what's the problem, again?

Tonio
Oh. Haha. I was thinking of setting to x->infinty instead of x->0. Stupid me.