How do I prove that the reciprocal of the limit of a variable is the same as the limit of the reciprocal of that same variable?

Question

If $$\lim_{p\to \infty}i^{(p)} = \delta$$

and we are computing

$$\lim_{p\to \infty} \frac{1}{i^{(p)}}$$

In what way can I justify the result to be $$\frac{1}{\delta}$$

Thank you.

Answer 1

As long as the limit is nonzero and no terms of the sequence are zero (or you are willing to ignore the ones that are) you can work directly from the $\epsilon-N$ definition. We are told that $\lim_{p \to \infty}i(p)=\delta$. That means that whatever $\epsilon \gt 0$ we choose we can find an $N$ so that $p \gt N \implies |i(p)-\delta| \lt \epsilon$

Now we claim that $\lim_{p \to \infty}\frac 1{i(p)}=\frac 1\delta$ If somebody gives us an $\epsilon' \gt 0$ we need $\left|\frac 1{i(p)}-\frac 1\delta \right| \lt \epsilon'$ or $\left|\frac{\delta-i(p)}{\delta i(p)}\right| \lt \epsilon'$. We will require that $i(p) \gt \frac 12\delta$, there is some $N_1$ that assures this. Then we need $\left|\frac{\delta-i(p)}{\delta i(p)}\right|\lt \left|\frac{\delta-i(p)}{\delta ^2}\right| \lt \epsilon'$ or $|\delta-i(p)| \lt \delta^2 \epsilon'$. There is some $N_2$ that assures this. Our $N$ is then the greater of $N_1,N_2$