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# Show that

Question:

Show that $\lim _{x \rightarrow 0} \frac{1}{|x|}=\infty$.

Solution:

Let $x=0+h$, when $x$ is tends to $0^{+}$

Since x tends to 0, h will also tend to 0.

Right Hand Limit(R.H.L):

$\lim _{x \rightarrow 0^{+}} f(x)$

$=\lim _{x \rightarrow 0^{+}} \frac{1}{|x|}$

$=\lim _{x \rightarrow 0^{+}} \frac{1}{(x)}$

$=\lim _{h \rightarrow 0^{+}} \frac{1}{(0+h)}$

$=\frac{1}{0}$

$=\infty$

Let $x=0-h$, when $x$ is tends to 0 -

ince $x$ tends to $0, \mathrm{~h}$ will also tend to 0 .

Left Hand Limit(L.H.L.):

$\lim _{x \rightarrow 0^{-}} f(x)$

$=\lim _{x \rightarrow 0^{-}} \frac{1}{|x|}$

$=\lim _{x \rightarrow 0^{-}} \frac{1}{(-x)}$

$=\lim _{h \rightarrow 0^{-}} \frac{1}{-(0-h)}$

$=\lim _{h \rightarrow 0^{-}} \frac{1}{h}$

$=\frac{1}{0}$

$=\infty$

Thus,

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$

$\therefore \lim _{x \rightarrow 0} \frac{1}{|x|}=\infty$