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  • Please solve RD Sharma class 12 chapter Continuity exercise 8.2 question 3 subquestion (vii) maths textbook solution

Please solve RD Sharma class 12 chapter Continuity exercise 8.2 question 3 subquestion (vii) maths textbook solution

Answers (1)

Answer:

 x = 0

Hint:

 Use L-Hospital Rule when we find 0/0 form

Given:

 f(x)=\left\{\begin{array}{cl} \frac{e^{x}-1}{\log _{e}(1-2 x)} & x \neq 0 \\ 7 & x=0 \end{array}\right.

Explanation:

At x = 0

\text { L.H.L } =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0}(0-h)=\lim _{h \rightarrow 0} \frac{e^{-h}-1}{\log _{e}(1-2 h)}

=\lim _{h \rightarrow 0} \frac{\frac{e^{-h}-1}{-h}}{\frac{\log _{e}(1-2 h)}{-2 h} \times 2}=\frac{1}{2}

\text { R.H.L }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}(0+h)=\lim _{h \rightarrow 0} \frac{e^{h}-1}{\log _{e}(1+2 h)}

=\lim _{h \rightarrow 0} \frac{\frac{e^{h}-1}{h}}{\frac{\log _{e}(2 h)}{2 h} \times 2}=\frac{1}{2}

\begin{aligned} &\text { Also } f(0)=7 \\ &\text { Thus L.H.L } =\text { R.H.L } \neq f(0) \end{aligned}

\thereforef(x) is not continuous at x = 0

Posted by

Gurleen Kaur

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