Show that

Question:

Show that $\lim _{x \rightarrow 0} \frac{1}{x}$ does not exist.

 

Solution:

Let $x=0+h$ for $x$ tending to $0^{+}$

Since x→ 0, h also tends to 0

Right Hand Limit(R.H.L.):

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

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

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

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

$=+\frac{1}{0}$

$=+\infty$

Let $x=0$-h for $x$ tending to $0^{-}$

Since $x \rightarrow 0$, h also tends to 0 .

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

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

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

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

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

$=-\frac{1}{0}$

$=-\infty$

Since,

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

 

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