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Question Number 165063 by MathsFan last updated on 25/Jan/22
limx→0xtanx−cosxxsinx−ex
Commented by MathsFan last updated on 25/Jan/22
anyhelp?
Answered by aleks041103 last updated on 27/Jan/22
limx→0xtanx−cosxxsinx−ex=Lfirstwesubstitute[00]−1[00]−1=?limx→0xtanx=elimtanx→0(x)ln(x)limx→0lnxctgx=l′hopitallimx→01x−1sin2x=−limx→0sin2xx==−limx→02sin(x)cos(x)1=0⇒limx→0xtan(x)=e0=1limx→0xsin(x)=elimlnx→0(x)sin(x)limx→0ln(x)sin(x)=limx→0ln(x)1sin(x)==limx→01x−cos(x)sin2x=−limx→0sin2xxcos(x)=−limx→0sin2xx=0⇒limxx→0sin(x)=e0=1⇒limx→0xtanx−cosxxsinx−ex=[1−11−1]=[00]tanx→xsinx→xl′hopital(xx)′=xx(xln(x))′=xx(1+ln(x))⇒L=limx→0xx(1+ln(x))+sin(x)xx(1+ln(x))−ex==limx→0xx(1+ln(x))xx(1+ln(x))−ex=1+limx→0exxx(1+ln(x))−ex==1+limx→01ln(x)=1L=1
Commented by aleks041103 last updated on 27/Jan/22
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