limite
Inviato: 21/12/2012, 16:05
qualcuno può correggere lo svolgimento di questo limite? grazie
\(\displaystyle \lim_{{x}\to 0} \frac{log(-3cos^2(x)+cos(x)+3)arccos(arcsin(x))}{ tg(x+45) tg\left(\frac{ 900 x^2}{ x+4 }\right)} \)
\(\displaystyle = \frac{0}{0}\)
\(\displaystyle 90 \lim_{{x}\to 0} \frac{ (x+4)(-3cos^2(x)+cos(x)+2) }{ 900 x^2 }\)
\(\displaystyle = \frac{1}{10} \lim_{{x}\to 0} \frac{-3cos^2(x)+cos(x)+2+ (x+4)(6sen(x)cos(x)-sen(x)) }{ 2x } \)
\(\displaystyle = \frac{1}{20} \lim_{{x}\to 0} [ 6sen(x)cos(x)-sen(x) + 6sen(x)cos(x)-sen(x)+ (x+4)(6cos^2(x)-6sen^2(x)-cos(x))] \)
\(\displaystyle = \frac{1}{20} [ 0-0 + 0-0+ (4)(6-0-1)] = 1\)
\(\displaystyle \lim_{{x}\to 0} \frac{log(-3cos^2(x)+cos(x)+3)arccos(arcsin(x))}{ tg(x+45) tg\left(\frac{ 900 x^2}{ x+4 }\right)} \)
\(\displaystyle = \frac{0}{0}\)
\(\displaystyle 90 \lim_{{x}\to 0} \frac{ (x+4)(-3cos^2(x)+cos(x)+2) }{ 900 x^2 }\)
\(\displaystyle = \frac{1}{10} \lim_{{x}\to 0} \frac{-3cos^2(x)+cos(x)+2+ (x+4)(6sen(x)cos(x)-sen(x)) }{ 2x } \)
\(\displaystyle = \frac{1}{20} \lim_{{x}\to 0} [ 6sen(x)cos(x)-sen(x) + 6sen(x)cos(x)-sen(x)+ (x+4)(6cos^2(x)-6sen^2(x)-cos(x))] \)
\(\displaystyle = \frac{1}{20} [ 0-0 + 0-0+ (4)(6-0-1)] = 1\)