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

Commented by fotosy2k last updated on 28/Aug/21

pls help

$${pls}\:{help} \\ $$

Answered by EDWIN88 last updated on 28/Aug/21

 lim_(x→0) (((e^(ax) −1)sin bx)/x^2 ) = lim_(x→0) ((e^(ax) −1)/x)×lim_(x→0) ((sin bx)/x)  = ab

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({e}^{{ax}} −\mathrm{1}\right)\mathrm{sin}\:{bx}}{{x}^{\mathrm{2}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{ax}} −\mathrm{1}}{{x}}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{bx}}{{x}} \\ $$$$=\:{ab} \\ $$

Commented by fotosy2k last updated on 30/Aug/21

thank you

$${thank}\:{you} \\ $$

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