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  • Find which of the functions is continuous or discontinuous at the indicated points: f(x) = e^1/x if x = 0

Find which of the functions is continuous or discontinuous at the indicated points:

f(x)=\left\{\begin{array}{cl} \frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}, \text { if } & x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right.

at x = 0

Answers (1)

Given,

f(x)=\left\{\begin{array}{cl} \frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}, \text { if } & x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right.

We need to check its continuity at x = 0
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-
\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)

Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 4 if -

\begin{aligned} &\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(0)\\ &\text { Clearly, }\\ &\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}\left\{\frac{e^{\frac{1}{-h}}}{1+e^{-\frac{1}{-h}}}\right\}_{\{u \operatorname{sing} \text { equation } 1\}}\\ &\Rightarrow \mathrm{LHL}=\frac{\mathrm{e}^{\frac{1}{-0}}}{1+\mathrm{e}^{-\frac{1}{-0}}}=\frac{\mathrm{e}^{-\infty}}{1+\mathrm{e}^{-\infty}}=\frac{0}{1+0}=0\\ &\therefore \mathrm{LHL}=0 . . .(2) \end{aligned}

Similarly, we proceed for RHL-

\begin{array}{l} \lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\left\{\frac{\mathrm{e}^{\frac{1}{\mathrm{~h}}}}{1+\mathrm{e}^{\frac{1}{\mathrm{~h}}}}\right\}_{\{\text {using equation } 1\}} \\ \quad \quad \lim _{\mathrm{h} \rightarrow 0}\left\{\frac{\mathrm{e}^{\frac{1}{\mathrm{~h}}}}{\mathrm{e}^{\frac{1}{\mathrm{~h}}\left(1+\mathrm{e}^{-\frac{1}{\mathrm{~h}}}\right)}}\right\}=\lim _{\mathrm{h} \rightarrow 0}\left\{\frac{1}{\left(1+\mathrm{e}^{-\frac{1}{\mathrm{~h}}}\right)}\right\} \end{array}

\begin{array}{l} \Rightarrow \mathrm{RHL}=\frac{1}{1+\mathrm{e}-0}=\frac{1}{1+\mathrm{e}^{-\infty}}=\frac{1}{1+0}=1 \\ \therefore \mathrm{RHL}=1 \ldots(3) \end{array}

And,
f(0) = 0 {using eqn 1} …(4)
 from equation 2 , 3 and 4 we can conclude that

\lim _{h \rightarrow 0} \mathrm{f}(0-\mathrm{h}) \neq \lim _{h \rightarrow 0} \mathrm{f}(0+\mathrm{h})

∴ f (x) is discontinuous at x = 0

Posted by

infoexpert22

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