The function \( f(x)=\begin{cases}\displaystyle\frac{e^{3x}-e^{-5x}}{x} & ,\ \text{if}\ x\ne 0\\\lambda & ,\ \text{if}\ x=0\end{cases} \) is continuous at \( x=0 \) for the value of ’\( \lambda \)’ as:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 8
Continuity at 0 needs \( \lambda=f(0)=\displaystyle\lim_{x\to 0}f(x) \). Splitting the limit and using the standard limit \( \lim_{x\to 0}\frac{e^{kx}-1}{x}=k \):
\( \displaystyle \lim_{x\to 0}\frac{e^{3x}-e^{-5x}}{x}=\lim_{x\to 0}\left[\frac{e^{3x}-1}{x}-\frac{e^{-5x}-1}{x}\right] \)
\( \displaystyle =3-(-5)=8 \)
So \( \lambda=8 \).