Question Number 173484 by mokys last updated on 12/Jul/22
![1) lim_(x→0) ((x^(sinx) −(sinx)^x )/(x^(cosx) +1)) 2) lim_(x→∞) [x lnx − 2x ln(sin(1/( (√x)))) ] how can solve this proplem ?](Q173484.png)
$$\left.\mathrm{1}\right)\:\boldsymbol{{lim}}_{\boldsymbol{{x}}\rightarrow\mathrm{0}} \:\frac{\boldsymbol{{x}}^{\boldsymbol{{sinx}}} −\left(\boldsymbol{{sinx}}\right)^{\boldsymbol{{x}}} }{\boldsymbol{{x}}^{\boldsymbol{{cosx}}} +\mathrm{1}} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\boldsymbol{{lim}}_{\boldsymbol{{x}}\rightarrow\infty} \:\left[\boldsymbol{{x}}\:\boldsymbol{{lnx}}\:−\:\mathrm{2}\boldsymbol{{x}}\:\boldsymbol{{ln}}\left(\boldsymbol{{sin}}\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{x}}}}\right)\:\right] \\ $$$$ \\ $$$$\boldsymbol{{how}}\:\boldsymbol{{can}}\:\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{proplem}}\:? \\ $$
Answered by aleks041103 last updated on 12/Jul/22
![1) lim_(x→0) sin(x)ln(x)= =lim_(x→0) ((ln(x))/(1/(sin(x))))=[(∞/∞)]=^(L′H) lim_(x→0) ((1/x)/((−cos(x))/(sin^2 (x))))= =−lim_(x→0) ((sin(x))/x) tg(x)=−1.0=0 ⇒lim_(x→0) ln(x^(sin(x)) )=0 ⇒ln(lim_(x→0) x^(sin(x)) )=0⇒lim_(x→0) x^(sin(x)) =1 lim_(x→0) x ln(sin(x))=lim_(x→0) ((ln(sin(x)))/(1/x))=[(∞/∞)]= =^(L′H) lim_(x→0) (((cos(x))/(sin(x)))/(−(1/x^2 )))=−((lim_(x→0) xcos(x))/(lim_(x→0) ((sin(x))/x)))=0 ⇒lim_(x→0) (sin(x))^x =1 1=0+1=0^1 +1=0^(cos(0)) +1=lim_(x→0) (x^(cos(x)) +1)=1 0=(0/1)=((1−1)/1)=(((lim_(x→0) x^(sin(x)) )−(lim_(x→0) (sin(x))^x ))/(lim_(x→0) (x^(cos(x)) +1))) ⇒lim_(x→0) ((x^(sin(x)) −(sin(x))^x )/(x^(cos(x)) +1))=0](Q173490.png)
$$\left.\mathrm{1}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{sin}\left({x}\right){ln}\left({x}\right)= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{ln}\left({x}\right)}{\frac{\mathrm{1}}{{sin}\left({x}\right)}}=\left[\frac{\infty}{\infty}\right]\overset{{L}'{H}} {=}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\frac{\mathrm{1}}{{x}}}{\frac{−{cos}\left({x}\right)}{{sin}^{\mathrm{2}} \left({x}\right)}}= \\ $$$$=−\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\left({x}\right)}{{x}}\:{tg}\left({x}\right)=−\mathrm{1}.\mathrm{0}=\mathrm{0} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{ln}\left({x}^{{sin}\left({x}\right)} \right)=\mathrm{0} \\ $$$$\Rightarrow{ln}\left(\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{x}^{{sin}\left({x}\right)} \right)=\mathrm{0}\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{x}^{{sin}\left({x}\right)} =\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{x}\:{ln}\left({sin}\left({x}\right)\right)=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{ln}\left({sin}\left({x}\right)\right)}{\frac{\mathrm{1}}{{x}}}=\left[\frac{\infty}{\infty}\right]= \\ $$$$\overset{{L}'{H}} {=}\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\frac{{cos}\left({x}\right)}{{sin}\left({x}\right)}}{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}=−\frac{\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{xcos}\left({x}\right)}{\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\left({x}\right)}{{x}}}=\mathrm{0} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\left({sin}\left({x}\right)\right)^{{x}} =\mathrm{1} \\ $$$$\mathrm{1}=\mathrm{0}+\mathrm{1}=\mathrm{0}^{\mathrm{1}} +\mathrm{1}=\mathrm{0}^{{cos}\left(\mathrm{0}\right)} +\mathrm{1}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\left({x}^{{cos}\left({x}\right)} +\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{0}=\frac{\mathrm{0}}{\mathrm{1}}=\frac{\mathrm{1}−\mathrm{1}}{\mathrm{1}}=\frac{\left(\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{x}^{{sin}\left({x}\right)} \right)−\left(\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left({sin}\left({x}\right)\right)^{{x}} \right)}{\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left({x}^{{cos}\left({x}\right)} +\mathrm{1}\right)} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}^{{sin}\left({x}\right)} −\left({sin}\left({x}\right)\right)^{{x}} }{{x}^{{cos}\left({x}\right)} +\mathrm{1}}=\mathrm{0} \\ $$
Answered by aleks041103 last updated on 12/Jul/22
![lim_(x→∞) [x lnx − 2x ln(sin((1/( (√x)))))]=L let z=(1/( (√x)))⇒x=(1/z^2 ) ⇒L=lim_(z→0) [((−2ln(z))/z^2 )−(2/z^2 )ln(sin(z))]= =−2 lim_(z→0) ((ln(z sin(z)))/z^2 )→−2((−∞)/0) ⇒L→+∞](Q173492.png)
$$\underset{{x}\rightarrow\infty} {{lim}}\left[{x}\:{lnx}\:−\:\mathrm{2}{x}\:{ln}\left({sin}\left(\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)\right)\right]={L} \\ $$$${let}\:{z}=\frac{\mathrm{1}}{\:\sqrt{{x}}}\Rightarrow{x}=\frac{\mathrm{1}}{{z}^{\mathrm{2}} } \\ $$$$\Rightarrow{L}=\underset{{z}\rightarrow\mathrm{0}} {{lim}}\left[\frac{−\mathrm{2}{ln}\left({z}\right)}{{z}^{\mathrm{2}} }−\frac{\mathrm{2}}{{z}^{\mathrm{2}} }{ln}\left({sin}\left({z}\right)\right)\right]= \\ $$$$=−\mathrm{2}\:\underset{{z}\rightarrow\mathrm{0}} {{lim}}\frac{{ln}\left({z}\:{sin}\left({z}\right)\right)}{{z}^{\mathrm{2}} }\rightarrow−\mathrm{2}\frac{−\infty}{\mathrm{0}} \\ $$$$\Rightarrow{L}\rightarrow+\infty \\ $$