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  • The function   <img alt="\mathrm{f(x)=\left\{\begin{array}{cc} \frac{c^{1 / x}-1,}{e^{1 / x}+1}, & x \neq 0 \\ 0, & x=0 \end{array}\right.}" src="https://entrancecorner.oncodecogs

The function   \mathrm{f(x)=\left\{\begin{array}{cc} \frac{c^{1 / x}-1,}{e^{1 / x}+1}, & x \neq 0 \\ 0, & x=0 \end{array}\right.}

Option: 1

is continuous at x=0


Option: 2

is not continuous at x=0


Option: 3

is not continuous at x=0, but can be made continuous at x=0


Option: 4

none of these.


Answers (1)

best_answer

\mathrm{\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0^{-}} f(-h)=\lim _{h \rightarrow 0} \frac{e^{-1 / h}-1}{e^{-1 / h}+1}=-1 . }
and,   \mathrm{\quad \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(h)=\lim _{x \rightarrow 0} \frac{e^{1 / h}-1}{e^{1 / h}+1} }
\mathrm{ =\lim _{h \rightarrow 0} \frac{1-e^{-1 / h}}{1+e^{-1 / h}}=1 \text {. } }

\mathrm{ \therefore \lim _{x \rightarrow 0^{-}} f(x), n f(x) }
Hence, f(x) is not continuous at x=0.

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avinash.dongre

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