Question

If \[ f(x)=\frac{(e^{2x}-1)\sin x^\circ}{x^2}, \quad x\neq0 \] is continuous at \(x=0\), then \(f(0)=\)

• A. \(\dfrac{90}{\pi}\)
• B. \(\dfrac{180}{\pi}\)
• C. \(\dfrac{\pi}{90}\)
• D. \(\dfrac{\pi}{180}\)

Hint:

Always convert degree–based trigonometric functions into radians while evaluating limits.

Answer 1

The Correct Option is C

Step 1: Use the definition of continuity.
For continuity at \(x=0\): \[ f(0)=\lim_{x\to0} f(x) \]
Step 2: Evaluate the limit.
\[ \lim_{x\to0}\frac{(e^{2x}-1)\sin x^\circ}{x^2} \]
Step 3: Convert degree to radian.
\[ \sin x^\circ = \sin\left(\frac{\pi x}{180}\right) \]
Step 4: Split the limit.
\[ = \lim_{x\to0}\frac{e^{2x}-1}{x}\cdot \lim_{x\to0}\frac{\sin\left(\frac{\pi x}{180}\right)}{x} \]
Step 5: Evaluate standard limits.
\[ \lim_{x\to0}\frac{e^{2x}-1}{x}=2 \] \[ \lim_{x\to0}\frac{\sin\left(\frac{\pi x}{180}\right)}{x} =\frac{\pi}{180} \]
Step 6: Final calculation.
\[ f(0)=2\cdot\frac{\pi}{180}=\frac{\pi}{90} \]