calculate lim_(x→0) ((ln(x+1+sin(πx)))/(xsin(2x)))


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Question Number 66304 by mathmax by abdo last updated on 12/Aug/19

$c a l c u l a t e {l i m}_{x \to 0} \frac{l n (x + 1 + s i n (π x))}{x s i n (2 x)}$
	Answered by kaivan.ahmadi last updated on 12/Aug/19
	$≡lim_(x→0) ((ln(x+1+πx))/(2x^2 )) =^(hop) lim_(x→0) ((1+π)/(4x(x+1+πx)))= { ((x→0^+ : ((1+π)/0^+ )=+∞)),((x→0^− : ((1+π)/0^− )=−∞)) :}$
	$\equiv {l i m}_{x \to 0} \frac{l n (x + 1 + π x)}{2 x^{2}} \overset{h o p}{=} {l i m}_{x \to 0} \frac{1 + π}{4 x (x + 1 + π x)} =$ ${\begin{cases} x \to 0^{+} : \frac{1 + π}{0^{+}} = + \infty \\ x \to 0^{-} : \frac{1 + π}{0^{-}} = - \infty \end{cases}$
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