Question

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

• A. Only I is true
• B. Only II is true
• C. Only Ill is true
• D. All are true

Answer 1

Answer: (D)

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

$x$	$1$	$0.3$	$0.2$	$0.1=10^{-1}$	$0.01=10^{-2}$	$0.001=10^{-3}$	$10^{-n}$
$f(x)$	$1$	$\mathrm{3.333......}$	$5$	$10$	$100=10^2$	$1000=10^3$	$10^n$

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
$\underset{x\to 0^{+}}{\lim }f(x)=+\infty$
(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

$x$	$-1$	$-0.3$	$-0.2$	$-10^{-1}$	$-10^{-2}$	$-10^{-3}$	$-10^{-n}$
$f(x)$	$-1$	$-\mathrm{3.333......}$	$-5$	$-10$	$-10^2$	$-10^3$	$-10^n$

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
$\underset{x\to 0^{-}}{\lim }f(x)=-\infty$

(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.