TS EAMCET · Maths · Limits
Define \(f: R \rightarrow R\) by \(f(x)= \begin{cases}(x-a) \frac{e^{\frac{1}{(x-a)}}-1}{\frac{1}{(x-a)}}, 1 & \text { for } x \neq a \ 0, & \text { at } x=a\end{cases}\) Then which one of the following is true?
- A Left and right limits of \(f\) at \(x=a\) are equal and they are not equal to \(f(a)\)
- B Both left and right limits of \(f\) at \(x=a\) exist and are not equal
- C The function \(f(x)\) is continuous at \(x=a\)
- D The function \(f(x)\) has a simple discontinuity at a point other than a
Answer & Solution
Correct Answer
(C) The function \(f(x)\) is continuous at \(x=a\)
Step-by-step Solution
Detailed explanation
We have, \(f(x)=\left\{\begin{array}{cc} (x-a) \frac{e^{\frac{1}{x-a}}-1}{\frac{1}{e^{x-a}}+1}, & x \neq a \\ 0, & x=a \end{array}\right.\)…
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