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limit ((e^tan3x)-1)/ln(1+sin2x) as x->0
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Is the limit : \[\Large \lim_{x\to0}\frac{e^{\tan 3x}-1}{\ln(1+\sin 2x)}\]??
yes
@amistre64 @ahoward79 @alf123 @precal @skullpatrol @supercrazy92 @sofiasores @Shianne95 @Shikanu @thomaster
use L-hospital
\[\Large \lim_{x\to0}\frac{e^{\tan 3x}-1}{\ln(1+\sin 2x)}\] Using L-Hospital, \[\Huge \lim_{x \rightarrow 0} \frac{e^{\tan3x} \times \frac{3}{1+9x^2}}{\frac{1}{1+\sin2x}\times \frac{2}{\sqrt{1-4x^2}}}\] Substitute x=0 now
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You should get 3/2 most probably?
If you don't get anything hit me up.
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