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

$${lim}_{x\to 0}\frac{x^{tanx}-cosx}{x^{sinx}-e^x}$$

Commented by MathsFan last updated on 25/Jan/22

$$anyhelp?$$

Answered by aleks041103 last updated on 27/Jan/22

$${lim}_{x\to 0}\frac{x^{tanx}-cosx}{x^{sinx}-e^x}=L$$$$firstwesubstitute$$$$\frac{\left[0^0\right]-1}{\left[0^0\right]-1}=?$$$$\underset{x\to 0}{lim}{x}^{tanx}=e^{\underset{x\to 0}{limtan}\left(x\right)ln\left(x\right)}$$$$\underset{x\to 0}{lim}\frac{lnx}{ctgx}\stackrel{l^{\prime }hopital}{=}\underset{x\to 0}{lim}\frac{\frac{1}{x}}{-\frac{1}{{sin}^{2}x}}=-\underset{x\to 0}{lim}\frac{{sin}^{2}x}{x}=$$$$=-\underset{x\to 0}{lim}\frac{2sin\left(x\right)cos\left(x\right)}{1}=0$$$$\Rightarrow \underset{x\to 0}{lim}{x}^{tan\left(x\right)}=e^0=1$$$$\underset{x\to 0}{lim}{x}^{sin\left(x\right)}=e^{\underset{x\to 0}{limln}\left(x\right)sin\left(x\right)}$$$$\underset{x\to 0}{lim}ln\left(x\right)sin\left(x\right)=\underset{x\to 0}{lim}\frac{ln\left(x\right)}{\frac{1}{sin\left(x\right)}}=$$$$=\underset{x\to 0}{lim}\frac{\frac{1}{x}}{-\frac{cos\left(x\right)}{{sin}^{2}x}}=-\underset{x\to 0}{lim}\frac{{sin}^{2}x}{xcos\left(x\right)}=-\underset{x\to 0}{lim}\frac{{sin}^{2}x}{x}=0$$$$\Rightarrow {\underset{x\to 0}{limx}}^{sin\left(x\right)}=e^0=1$$$$\Rightarrow {lim}_{x\to 0}\frac{x^{tanx}-cosx}{x^{sinx}-e^x}=\left[\frac{1-1}{1-1}\right]=\left[\frac{0}{0}\right]$$$$tanx\to x$$$$sinx\to x$$$$l^{\prime }hopital$$$${\left(x^x\right)}^{\prime }=x^x{\left(xln\left(x\right)\right)}^{\prime }=x^x(1+ln\left(x\right))$$$$\Rightarrow L=\underset{x\to 0}{lim}\frac{x^x(1+ln\left(x\right))+sin\left(x\right)}{x^x(1+ln\left(x\right))-e^x}=$$$$=\underset{x\to 0}{lim}\frac{x^x(1+ln\left(x\right))}{x^x(1+ln\left(x\right))-e^x}=1+\underset{x\to 0}{lim}\frac{e^x}{x^x(1+ln\left(x\right))-e^x}=$$$$=1+\underset{x\to 0}{lim}\frac{1}{ln\left(x\right)}=1$$$$L=1$$

Commented by aleks041103 last updated on 27/Jan/22

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