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The function f(x)={e3x−e−5xx,x≠0λ,x=0f(x)=\begin{cases}\frac{e^{3x}-e^{-5x}}{x}, & x\neq 0\\ \lambda, & x=0\end{cases} is continuous at x=0 for the value of 'λ\lambda' as:

Solution

✅ Correct Option: 2

Split as e3x−1x−e−5x−1x\frac{e^{3x}-1}{x}-\frac{e^{-5x}-1}{x}. Using lim⁡x→0ekx−1x=k\lim_{x\to0}\frac{e^{kx}-1}{x}=k, the limit is 3−(−5)=83-(-5)=8. For continuity λ\lambda must equal this limit, so λ=8\lambda=8.

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