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Question Number 170809 by thean last updated on 31/May/22

Answered by som(math1967) last updated on 31/May/22

$$\underset{x\to 0}{lim}\frac{e^{-1}(e^{2x}-1)sin3x}{x^2}$$$$=\frac{6}{e}\underset{x\to 0}{lim}\frac{(e^{2x}-1)}{2x}\times \frac{sin3x}{3x}$$$$=\frac{6}{e}\times 1=\frac{6}{e}$$

Answered by Mathspace last updated on 31/May/22

$$f\left(x\right)=\frac{e^{2x-1}-e^{-1}}{x^2}sin\left(3x\right)$$$$sin\left(3x\right)\sim 3xand$$$$e^{2x-1}-e^{-1}=e^{-1}(e^{2x}-1)$$$$e^u\sim 1+u\Rightarrow e^{2x}-1\sim 2x\Rightarrow$$$$f\left(x\right)\sim \frac{2{xe}^{-1}}{x^2}\times 3x=\frac{6}{e}$$$$\Rightarrow {lim}_{x\to 0}f\left(x\right)=\frac{6}{e}$$

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