Question

Which of the following function is discontinuous at \(x = 0\) ?

• A. \(f(x) = (1+x)^{\frac{2}{x}},\) for \(x\neq 0\)
  \(= e^2,\) for \(x = 0\)
• B. \(f(x) = sinx-cosx,\) for \(x\neq 0\)
  \(= -1,\) for \(x = 0\)
• C. \(f(x) = \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1},\) for \(x\neq 0\)
  \(= -1,\) for \(x = 0\)
• D. \(f(x) = \frac{e^{5x}-e^{2x}}{sin3x},\) for \(x\neq 0\)
  \(= 1,\) for \(x = 0\)

Hint:

Compare left and right limits at \(x=0\) for each piece.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
A function is continuous at \(0\) if the left limit, right limit and value are all equal.

Step 2: Key Formula or Approach
Compute each limit using standard results.

Step 3: Detailed Explanation
(A) \((1+x)^{2/x}\to e^2=f(0)\): continuous.
(B) \(\sin x-\cos x\to-1=f(0)\): continuous.
(D) \(\dfrac{e^{5x}-e^{2x}}{\sin3x}=\dfrac{e^{2x}(e^{3x}-1)}{\sin3x}\to1\cdot1=1=f(0)\): continuous.
(C) As \(x\to0^+\), \(e^{1/x}\to\infty\) so the ratio tends to \(1\). As \(x\to0^-\), \(e^{1/x}\to0\) so the ratio tends to \(-1\). The two limits differ, so \(f\) is discontinuous at \(0\).

Final Answer:
Function (C) is discontinuous at \(x=0\), option (C). \[ \boxed{\text{(C)}} \]