The limit of $1/x$ as $x\to 0$.

Question

I need help with the following limit. I know that the answer is positive infinity but could someone break this down to me step by step please. I got confused since I intuitionally tried to plug in $x=0$ but then we can't divide by zero! Also we did not yet cover the Hopital rule so I guess I am expected to solve in some other way. Thank you.

Answer 1

To best understand the behavior of $\frac1x$ as $x$ gets arbitrarily close to $0$, it helps to look at a graph:

Of course, you're probably already familiar with this, so another thing that will help is a table of values as $x$ gets small: $$\begin{array}{|c|c|} \hline x & \frac1x \\ \hline 1 & 1 \\ \hline 0.1 & 10 \\ \hline 0.01 & 100 \\ \hline 0.001 & 1000 \\ \hline \end{array} \ \ \ \begin{array}{|c|c|} \hline x & \frac1x \\ \hline -1 & -1 \\ \hline -0.1 & -10 \\ \hline -0.01 & -100 \\ \hline -0.001 & -1000 \\ \hline \end{array}$$

Notice how these values seem to diverge: this should give you a correct intuition as to how this limit works as $x\to0$.

In this case, the "right limit" (approaching $x=0$ from the first table above) diverges to positive infinity while the "left limit" (the second table) diverges to negative infinity. It should make sense then that $\left(\lim_{x\to0}\frac1x\right)$ is undefined: for it to be defined, it must have left and right limits that are equal to the same real number.

Answer 2

$\lim_{x \rightarrow 0^{+}} f(x) = +\infty$ means that for $\forall M >0$ $\exists \sigma: 0<x<\sigma \Rightarrow f(x) > M$

Now plug in your function into this definition.