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Question Number 107828 by Rohit@Thakur last updated on 12/Aug/20

lim {((x^2 +5x+3)/(x^2 +x+2))}^x x→0 lim ((10^x −2^x −5^x +1)/(xtanx)) x→0

$${lim}\:\:\:\:\:\:\left\{\frac{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{3}}{{x}^{\mathrm{2}} +{x}+\mathrm{2}}\right\}^{{x}} \\ $$$${x}\rightarrow\mathrm{0} \\ $$$${lim}\:\:\:\:\frac{\mathrm{10}^{{x}} −\mathrm{2}^{{x}} −\mathrm{5}^{{x}} +\mathrm{1}}{{xtanx}} \\ $$$${x}\rightarrow\mathrm{0} \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 12/Aug/20

lim_(x→0) (((5^x −1)(2^x −1))/(xsinx)).cosx lim_(x→0) (((5^x −1)(2^x −1))/(x.x))cosx (sinx→x) lim_(x→0) ((5^x −1)/x).((2^x −1)/x)=log5log2

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{5}^{{x}} −\mathrm{1}\right)\left(\mathrm{2}^{{x}} −\mathrm{1}\right)}{{xsinx}}.{cosx}\:\:\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{5}^{{x}} −\mathrm{1}\right)\left(\mathrm{2}^{{x}} −\mathrm{1}\right)}{{x}.{x}}{cosx}\:\:\:\:\left({sinx}\rightarrow{x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{5}^{{x}} −\mathrm{1}}{{x}}.\frac{\mathrm{2}^{{x}} −\mathrm{1}}{{x}}={log}\mathrm{5}{log}\mathrm{2} \\ $$

Answered by mathmax by abdo last updated on 13/Aug/20

f(x) ={((x^2 +5x +3)/(x^2 +x+2))}^x ⇒f(x) =e^(xln(((x^2 +5x+3)/(x^2 +x+2)))) ⇒f is continue at 0 and lim_(x→0) f(x) =e^0 =1

$$\mathrm{f}\left(\mathrm{x}\right)\:=\left\{\frac{\mathrm{x}^{\mathrm{2}} \:+\mathrm{5x}\:+\mathrm{3}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{2}}\right\}^{\mathrm{x}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{\mathrm{xln}\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{3}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}+\mathrm{2}}\right)} \:\Rightarrow\mathrm{f}\:\mathrm{is}\:\mathrm{continue}\:\mathrm{at}\:\mathrm{0}\: \\ $$$$\mathrm{and}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{\mathrm{0}} \:=\mathrm{1} \\ $$

Answered by mathmax by abdo last updated on 13/Aug/20

g(x) =((10^x −2^x −5^x +1)/(xtanx)) ⇒g(x) =((e^(xln(10)) −e^(xln(2)) −e^(xln(5)) +1)/(xtanx)) let use hospital theorem ⇒ lim_(x→0) g(x) =lim_(x→0) ((ln(10)e^(xln10) −ln(2)e^(2ln2) −ln5 e^(xln5) )/(tanx +x(1+tan^2 x))) =lim_(x→0) ((ln^2 (10)e^(xln10) −ln^2 (2)e^(2ln2) −ln^2 5 e^(xln5) )/(1+tan^2 x +1+tan^2 x +x(....))) =((ln^2 (10)−ln^2 (2)−ln^2 5)/2) =(((ln(2)+ln5)^2 −ln^2 2−ln^2 5)/2) =((2ln(10))/2) =ln(10)

$$\mathrm{g}\left(\mathrm{x}\right)\:=\frac{\mathrm{10}^{\mathrm{x}} −\mathrm{2}^{\mathrm{x}} −\mathrm{5}^{\mathrm{x}} +\mathrm{1}}{\mathrm{xtanx}}\:\Rightarrow\mathrm{g}\left(\mathrm{x}\right)\:=\frac{\mathrm{e}^{\mathrm{xln}\left(\mathrm{10}\right)} −\mathrm{e}^{\mathrm{xln}\left(\mathrm{2}\right)} −\mathrm{e}^{\mathrm{xln}\left(\mathrm{5}\right)} +\mathrm{1}}{\mathrm{xtanx}} \\ $$$$\mathrm{let}\:\mathrm{use}\:\mathrm{hospital}\:\mathrm{theorem}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{g}\left(\mathrm{x}\right)\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{ln}\left(\mathrm{10}\right)\mathrm{e}^{\mathrm{xln10}} −\mathrm{ln}\left(\mathrm{2}\right)\mathrm{e}^{\mathrm{2ln2}} −\mathrm{ln5}\:\mathrm{e}^{\mathrm{xln5}} }{\mathrm{tanx}\:\:+\mathrm{x}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{10}\right)\mathrm{e}^{\mathrm{xln10}} −\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\mathrm{e}^{\mathrm{2ln2}} −\mathrm{ln}^{\mathrm{2}} \mathrm{5}\:\mathrm{e}^{\mathrm{xln5}} }{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}\:\:+\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}\:+\mathrm{x}\left(....\right)} \\ $$$$=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{10}\right)−\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{ln}^{\mathrm{2}} \mathrm{5}}{\mathrm{2}}\:=\frac{\left(\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{ln5}\right)^{\mathrm{2}} −\mathrm{ln}^{\mathrm{2}} \mathrm{2}−\mathrm{ln}^{\mathrm{2}} \mathrm{5}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{2ln}\left(\mathrm{10}\right)}{\mathrm{2}}\:=\mathrm{ln}\left(\mathrm{10}\right) \\ $$

Commented by Dwaipayan Shikari last updated on 13/Aug/20

Commented by Dwaipayan Shikari last updated on 13/Aug/20

How sir?

$${How}\:{sir}? \\ $$

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