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Question Number 152433 by fotosy2k last updated on 28/Aug/21

Commented by fotosy2k last updated on 28/Aug/21

$$plshelp$$

Answered by EDWIN88 last updated on 28/Aug/21

$$\underset{x\to 0}{\lim }\frac{(e^{ax}-1)\sin bx}{x^2}=\underset{x\to 0}{\lim }\frac{e^{ax}-1}{x}\times \underset{x\to 0}{\lim }\frac{\sin bx}{x}$$$$=ab$$

Commented by fotosy2k last updated on 30/Aug/21

$$thankyou$$

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