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Q. If $f(x)=\frac{1}{x}$, then which of the following statements is /are true?
I. $\displaystyle\lim _{x \rightarrow 0^{+}} f(x)=+\infty$
II. $\displaystyle\lim _{x \rightarrow 0^{-}} f(x)=-\infty$
III. $\displaystyle\lim _{x \rightarrow 0} f(x)$ does not exist.

Continuity and Differentiability

Solution:

To analyse the function $f(x)=\frac{1}{x}$ near $x=0$. We follow the usual trick of finding the value of the function at real numbers close to 0 . Essentially we are trying to find the right hand limit of $f$ at 0 . We tabulate this in the following table
$x$ $1$ $0.3$ $0.2$ $0.1 = 10^{-1}$ $0.01 = 10^{-2}$ $0.001 = 10^{-3}$ $10^{-n}$
$f(x)$ $1$ $3.333......$ $5$ $10$ $100 = 10^2$ $1000 = 10^3$ $10^n$
We observe that as $x$ gets closer to 0 from the right, the value of $f(x)$ shoots up higher. This may be rephrased as the value of $f(x)$ may be made larger than any given number by choosing a positive real number very close to 0 .
In symbols, we write
$\displaystyle\lim _{x \rightarrow 0^{+}} f(x)=+\infty$
(to be read as the right hand limit of $f(x)$ at 0 is plus infinity). We wish to emphasise that $+\infty$ is not a real number and hence the right hand limit of $f$ at 0 does not exist (as a real number).
Similarly, the left hand limit of $f$ at 0 may be found. The following table is self explanatory
$x$ $-1$ $-0.3$ $-0.2$ $- 10^{-1}$ $-10^{-2}$ $- 10^{-3}$ $-10^{-n}$
$f(x)$ $-1$ $-3.333......$ $-5$ $-10$ $- 10^2$ $- 10^3$ $-10^n$

From the Table II, we deduce that the value of $f(x)$ may be made smaller than any given number by choosing a negative real number very close to 0 . In symbols, we write
$\displaystyle\lim _{x \rightarrow 0^{-}} f(x)=-\infty$
image
(to be read as the left hand limit of $f(x)$ at 0 is $-\infty$ ). Again, we wish to emphasise that $-\infty$ is not a real number and hence the left hand limit of $f$ at 0 does not exist (as a real number). The graph of the reciprocal function given in figure is a geometric representation of the above mentioned facts.