Question Number 73555 by Rio Michael last updated on 13/Nov/19
![find a) Lim_(x→−∞) ((ln(1+e^x ))/e^x ) b) lim_(x→+∞) ((√(x^2 + 3x)) −x) c) lim_(x→0) (√x) ln(sinx) d) lim_(x→+∞) ((e^(2x+1) −e^x )/(x^2 −x−1)) e) lim_(x→−∞) [(√(1−xe^x )) ]](Q73555.png)
$${find}\: \\ $$$$\left.{a}\right)\:\:\underset{{x}\rightarrow−\infty} {{Lim}}\:\frac{{ln}\left(\mathrm{1}+{e}^{{x}} \right)}{{e}^{{x}} } \\ $$$$\left.{b}\right)\:\underset{{x}\rightarrow+\infty} {{lim}}\:\left(\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}}\:\:−{x}\right) \\ $$$$\left.{c}\right)\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\sqrt{{x}}\:{ln}\left({sinx}\right) \\ $$$$\left.{d}\right)\:\:\underset{{x}\rightarrow+\infty} {{lim}}\:\frac{{e}^{\mathrm{2}{x}+\mathrm{1}} −{e}^{{x}} }{{x}^{\mathrm{2}} −{x}−\mathrm{1}} \\ $$$$\left.{e}\right)\:\underset{{x}\rightarrow−\infty} {{lim}}\left[\sqrt{\mathrm{1}−{xe}^{{x}} \:}\:\right] \\ $$
Commented by mathmax by abdo last updated on 13/Nov/19
$$\left.{a}\right)\:{we}\:{use}\:{chang}.\:{e}^{{x}} ={t}\:\Rightarrow{lim}_{{x}\rightarrow−\infty} \frac{{ln}\left(\mathrm{1}+{e}^{{x}} \right)}{{e}^{{x}} }\:={lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}} \\ $$$$=\mathrm{1} \\ $$$$\left.{b}\right)\:{we}\:{have}\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}{x}}−{x}\:=\sqrt{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{3}}{{x}}\right)}\:−{x}\:\: \\ $$$$={x}\sqrt{\mathrm{1}+\frac{\mathrm{3}}{{x}}}−{x}\:\sim{x}\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}{x}}\right)−{x}\:\:\:\left({x}\rightarrow+\infty\right)\: \\ $$$$\sim\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} \sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}−{x}\:=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left.{c}\right)\:{lim}_{{x}\rightarrow\mathrm{0}} \sqrt{{x}}{ln}\left({sinx}\right)?\:\:\:\:{we}\:{have}\:{sinx}\:\sim{x}\:\Rightarrow\sqrt{{x}}{ln}\left({sinx}\right)\sim\sqrt{{x}}{ln}\left({x}\right) \\ $$$${and}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\sqrt{{x}}{ln}\left({x}\right)\:=_{\sqrt{{x}}={u}} \:\:{lim}_{{u}\rightarrow\mathrm{0}} \:\:{uln}\left({u}^{\mathrm{2}} \right)\:={lim}_{{u}\rightarrow\mathrm{0}} \mathrm{2}{uln}\left({u}\right)=\mathrm{0} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 13/Nov/19
$$\left.{d}\right)\:{let}\:{f}\left({x}\right)\:=\frac{{e}^{\mathrm{2}{x}+\mathrm{1}} −{e}^{{x}} }{{x}^{\mathrm{2}} −{x}−\mathrm{1}}\:\Rightarrow{f}\left({x}\right)=\frac{{e}^{\mathrm{2}{x}+\mathrm{1}} }{{x}^{\mathrm{2}} }×\frac{\mathrm{1}−{e}^{{x}−\mathrm{2}{x}−\mathrm{1}} }{\mathrm{1}−\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$${we}\:{have}\:{lim}_{{x}\rightarrow+\infty} \:\:\:\frac{{e}^{\mathrm{2}{x}+\mathrm{1}} }{{x}^{\mathrm{2}} }\:=+\infty\:\:\:\left({e}^{{x}} \:{defeat}\:{x}^{{n}} \:...\right) \\ $$$${lim}_{{x}\rightarrow+\infty} \:\:\:\frac{\mathrm{1}−{e}^{−{x}−\mathrm{1}} }{\mathrm{1}−\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\:=\mathrm{1}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=+\infty \\ $$
Commented by Rio Michael last updated on 14/Nov/19
$${thanks}\:{sir} \\ $$
Commented by malwaan last updated on 14/Nov/19
![b) lim_(x→∞) ((((√(x^2 +3x))−x)((√(x^2 +3x))+x))/((√(x^2 +3x))+x)) = lim_(x→∞) ((x^2 +3x−x^2 )/(x[(√(1+(3/x)))+1])) =lim_(x→∞) (3/((√(1+(3/x)))+1))= (3/2)](Q73640.png)
$$\left.\boldsymbol{{b}}\right)\:\underset{\boldsymbol{{x}}\rightarrow\infty} {\boldsymbol{{lim}}}\frac{\left(\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}}−\boldsymbol{{x}}\right)\left(\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}}+\boldsymbol{{x}}\right)}{\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}}+\boldsymbol{{x}}}\: \\ $$$$=\:\underset{\boldsymbol{{x}}\rightarrow\infty} {\boldsymbol{{lim}}}\frac{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}−\boldsymbol{{x}}^{\mathrm{2}} }{\boldsymbol{{x}}\left[\sqrt{\mathrm{1}+\frac{\mathrm{3}}{\boldsymbol{{x}}}}+\mathrm{1}\right]}\: \\ $$$$=\underset{\boldsymbol{{x}}\rightarrow\infty} {\boldsymbol{{lim}}}\frac{\mathrm{3}}{\sqrt{\mathrm{1}+\frac{\mathrm{3}}{\boldsymbol{{x}}}}+\mathrm{1}}=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by ajfour last updated on 13/Nov/19
$$\left.{a}\right)\:\:\:{e}^{−\infty} \rightarrow\mathrm{0}\:\:\:{and}\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{1}+{h}\right)}{{h}}=\mathrm{1} \\ $$$${Ans}.\:{to}\:\left({a}\right)\:=\:\mathrm{1} \\ $$$$\left.{b}\right)\:\underset{{x}\rightarrow+\infty} {{lim}}\:\left(\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}}\:\:−{x}\right) \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}{x}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}+{x}}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}}{\sqrt{\mathrm{1}+\frac{\mathrm{3}}{{x}}}+\mathrm{1}} \\ $$$$\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\: \\ $$$$\left.{c}\right)\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\sqrt{{x}}\:{ln}\left({sinx}\right) \\ $$$$=\frac{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{d}}{{dx}}\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)}{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{\sqrt{{x}}}\right)}=\frac{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{tan}\:{x}}}{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(−\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{{x}}}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{x}}{\mathrm{tan}\:{x}}\right)×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(−\mathrm{2}\sqrt{{x}}\right) \\ $$$$=\:\mathrm{1}×\mathrm{0}\:=\:\mathrm{0}\:. \\ $$$$\left.{d}\right)\:\:\underset{{x}\rightarrow+\infty} {{lim}}\:\frac{{e}^{\mathrm{2}{x}+\mathrm{1}} −{e}^{{x}} }{{x}^{\mathrm{2}} −{x}−\mathrm{1}} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}{e}^{{x}} \left(\frac{{e}×{e}^{{x}} −\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\right) \\ $$$$\:\:\:\:\:=\:\infty\:. \\ $$
Commented by Rio Michael last updated on 13/Nov/19
$${thanks}\:{sir} \\ $$