If $\lim\limits_{x \to \infty} f(x) = A$ then prove $\lim\limits_{x \to \infty} \frac{1}{f(x)} =\frac{1}{A}$ This is a part of my homework and I was asked to prove that if $\lim\limits_{x\to \infty} f(x) = A$, then $\lim\limits_{x\to \infty} \frac{1}{f(x)} = \frac{1}{A}$.

In the previous question, I was asked to show for $\lim\limits_{x\to \infty} f(x) = A$, and $\lim\limits_{x\to \infty} g(x) = B$, $\lim\limits_{x\to \infty} f(x)+g(x) = A+B$

I solved that by showing $|f(x)+g(x) - (A+B)| \le |f(x)-A| + |g(x)-B| = \frac{\epsilon}{2} + \frac{\epsilon}{2}$ but I am not sure if this help on this question
Assume $A\neq0$.

Set $\epsilon&gt;0$. We want $\left|\dfrac1{f(x)}-\dfrac1{A}\right|&lt;\epsilon$ by making $|f(x)-A|$ as small as we please.

Now, $\left|\dfrac1{f(x)}-\dfrac1{A}\right|&lt;\epsilon\implies \left|\dfrac{f(x)-A}{A\cdot f(x)}\right|&lt;\epsilon$.

We can make $|f(x)-A|$ as small as we like, so that we only need to worry about $\left|\dfrac1{A\cdot f(x)}\right|$.

To bound $|A\cdot f(x)|$, make $|f(x)-A|&lt;\dfrac{|A|}2$ which puts a lower bound on $|f(x)|$ of $\dfrac{|A|}{2}$.

Thus, $\left|\dfrac1{A\cdot f(x)}\right|&lt;\dfrac1{|A|\cdot \left|\dfrac{A}{2}\right|}=\dfrac{2}{|A|^2}$. 

Now, we can write this down formally.

Set $\epsilon&gt;0$.

Since $\lim_{x\to\infty} f(x) = A$, there is some $N_0$ such that $|f(x)-A|&lt;\min(\dfrac{|A|^2\epsilon}2,\dfrac{|A|}{2})$ for all $x&gt;N_0$. 

So, suppose $x&gt;N_0$. Since, $|f(x)-A|&lt;\dfrac {|A|}2$, $\dfrac{|A|}{2}&lt;|f(x)|$.

Thus, $\left|\dfrac1{f(x)}-\dfrac1{A}\right|= \left|\dfrac{f(x)-A}{A\cdot f(x)}\right|=\dfrac{|f(x)-A|}{|A|\cdot |f(x)|}&lt;\dfrac{\dfrac{{|A|}^2}{2}\epsilon}{|A|\cdot \dfrac{|A|}{2}}=\epsilon$.
The statement is only true if $A \neq 0$. If you believe that $\lim_{x\to\infty}\frac1x =0$, then put $g(x)=1/x$ and note that $\lim 1/f(x)=\lim g(f(x))= g(\lim f(x)) = g(A)=1/A$ since $g$ is continuous at $A\neq 0$.