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Question Number 165063 by MathsFan last updated on 25/Jan/22

 lim_(x→0) ((x^(tanx) −cosx)/(x^(sinx) −e^x ))

limx0xtanxcosxxsinxex

Commented by MathsFan last updated on 25/Jan/22

any help?

anyhelp?

Answered by aleks041103 last updated on 27/Jan/22

lim_(x→0) ((x^(tanx) −cosx)/(x^(sinx) −e^x ))=L  first we substitute  (([0^0 ]−1)/([0^0 ]−1))=?  lim_(x→0)  x^(tanx) =e^(lim_(x→0) tan(x)ln(x))   lim_(x→0)  ((lnx)/(ctgx))=^(l′hopital)  lim_(x→0) ((1/x)/(−(1/(sin^2 x))))=−lim_(x→0) ((sin^2 x)/x)=  =−lim_(x→0) ((2sin(x)cos(x))/1)=0  ⇒lim_(x→0)  x^(tan(x)) =e^0 =1  lim_(x→0)  x^(sin(x)) =e^(lim_(x→0) ln(x) sin (x))   lim_(x→0)  ln(x)sin(x)=lim_(x→0)  ((ln(x))/(1/(sin(x))))=  =lim_(x→0) ((1/x)/(−((cos(x))/(sin^2 x))))=−lim_(x→0) ((sin^2 x)/(x cos(x)))=−lim_(x→0) ((sin^2 x)/x)=0  ⇒lim_(x→0) x^(sin(x)) =e^0 =1  ⇒lim_(x→0) ((x^(tanx) −cosx)/(x^(sinx) −e^x ))=[((1−1)/(1−1))]=[(0/0)]  tanx→x  sinx→x  l′hopital  (x^x )′=x^x (x ln(x))′=x^x (1+ln(x))  ⇒L=lim_(x→0) ((x^x (1+ln(x))+sin(x))/(x^x (1+ln(x))−e^x ))=  =lim_(x→0) ((x^x (1+ln(x)))/(x^x (1+ln(x))−e^x ))=1+lim_(x→0) (e^x /(x^x (1+ln(x))−e^x ))=  =1+lim_(x→0) (1/(ln(x)))=1  L=1

limx0xtanxcosxxsinxex=Lfirstwesubstitute[00]1[00]1=?limx0xtanx=elimtanx0(x)ln(x)limx0lnxctgx=lhopitallimx01x1sin2x=limx0sin2xx==limx02sin(x)cos(x)1=0limx0xtan(x)=e0=1limx0xsin(x)=elimlnx0(x)sin(x)limx0ln(x)sin(x)=limx0ln(x)1sin(x)==limx01xcos(x)sin2x=limx0sin2xxcos(x)=limx0sin2xx=0limxx0sin(x)=e0=1limx0xtanxcosxxsinxex=[1111]=[00]tanxxsinxxlhopital(xx)=xx(xln(x))=xx(1+ln(x))L=limx0xx(1+ln(x))+sin(x)xx(1+ln(x))ex==limx0xx(1+ln(x))xx(1+ln(x))ex=1+limx0exxx(1+ln(x))ex==1+limx01ln(x)=1L=1

Commented by aleks041103 last updated on 27/Jan/22

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