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Question-150567

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Question Number 150567 by DELETED last updated on 13/Aug/21
Answered by DELETED last updated on 13/Aug/21
1). lim_(x→∞) ((3x^2 −4)/(2x^2 +x+1)) =lim_(x→∞) ((3x^2 /x^2 −4/x^2 )/(2x^2 /x^2 +x/x^2 +1/x^2 )) = lim_(x→∞) ((3−4/x^2 )/(2+1/x+1/x^2 )) =((3−4/∞^2 )/(2+1/∞+1/∞^2 )) =((3−0)/(2+0+0))=(3/2) //
$$\left.\mathrm{1}\right).\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3x}^{\mathrm{2}} −\mathrm{4}}{\mathrm{2x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3x}^{\mathrm{2}} /\mathrm{x}^{\mathrm{2}} −\mathrm{4}/\mathrm{x}^{\mathrm{2}} }{\mathrm{2x}^{\mathrm{2}} /\mathrm{x}^{\mathrm{2}} +\mathrm{x}/\mathrm{x}^{\mathrm{2}} +\mathrm{1}/\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3}−\mathrm{4}/\mathrm{x}^{\mathrm{2}} }{\mathrm{2}+\mathrm{1}/\mathrm{x}+\mathrm{1}/\mathrm{x}^{\mathrm{2}} }\:=\frac{\mathrm{3}−\mathrm{4}/\infty^{\mathrm{2}} }{\mathrm{2}+\mathrm{1}/\infty+\mathrm{1}/\infty^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\mathrm{3}−\mathrm{0}}{\mathrm{2}+\mathrm{0}+\mathrm{0}}=\frac{\mathrm{3}}{\mathrm{2}}\:// \\ $$
Answered by DELETED last updated on 13/Aug/21
2). lim_(x→∞) ((4x^5 +6x^2 −7)/(x^5 +7x+3))=lim_(x→∞) ((((4x^5 )/x^5 )+((6x^2 )/x^5 ) −(7/x^5 ))/((x^5 /x^5 ) + ((7x)/x^5 ) +(3/x^5 ))) =lim_(x→∞) ((4+(6/x^3 )−(7/x^5 ))/(1+(7/x^4 )+(3/x^5 ))) = ((4+(6/∞^3 ) −(7/∞^5 ))/(1+(7/∞^4 )+(3/∞^5 ))) =((4+0−0)/(1+0+0))=4//
$$\left.\mathrm{2}\right).\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{4x}^{\mathrm{5}} +\mathrm{6x}^{\mathrm{2}} −\mathrm{7}}{\mathrm{x}^{\mathrm{5}} +\mathrm{7x}+\mathrm{3}}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\frac{\mathrm{4x}^{\mathrm{5}} }{\mathrm{x}^{\mathrm{5}} }+\frac{\mathrm{6x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{5}} }\:−\frac{\mathrm{7}}{\mathrm{x}^{\mathrm{5}} }}{\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{x}^{\mathrm{5}} }\:+\:\frac{\mathrm{7x}}{\mathrm{x}^{\mathrm{5}} }\:+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{5}} }} \\ $$$$\:\:\:\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{4}+\frac{\mathrm{6}}{\mathrm{x}^{\mathrm{3}} }−\frac{\mathrm{7}}{\mathrm{x}^{\mathrm{5}} }}{\mathrm{1}+\frac{\mathrm{7}}{\mathrm{x}^{\mathrm{4}} }+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{5}} }}\:\:=\:\frac{\mathrm{4}+\frac{\mathrm{6}}{\infty^{\mathrm{3}} }\:−\frac{\mathrm{7}}{\infty^{\mathrm{5}} }}{\mathrm{1}+\frac{\mathrm{7}}{\infty^{\mathrm{4}} }+\frac{\mathrm{3}}{\infty^{\mathrm{5}} }} \\ $$$$\:\:=\frac{\mathrm{4}+\mathrm{0}−\mathrm{0}}{\mathrm{1}+\mathrm{0}+\mathrm{0}}=\mathrm{4}// \\ $$
Answered by DELETED last updated on 13/Aug/21
3). lim_(x→∞) ((√(5x+2)) −(√(3x−1)) )×(((√(5x+2)) +(√(3x−1)) ))/( (√(5x+2)) ×(√(3x−1)) ))) =lim_(x→∞) (((5x+2)−(3x−1))/( (√(5x+2)) +(√(3x−1)) ))) =lim_(x→∞) ((2x+3)/( (√(5x+2)) +(√(3x−1)) ))) =lim_(x→∞) ((2x/(√x)+3/(√x))/( (√(5x/x+2/x)) +(√(3x/x−1/x)) ))) =lim_(x→∞) ((2(√x)+(3/( (√x))))/( (√(5+(2/x))) +(√(3−(1/x))) ))) = ((2(√∞)+(3/( (√∞))))/( (√(5+(2/∞))) +(√(3−(1/∞))) ))) =((∞+0)/( (√(5+0)) +(√(3−0)))) =(∞/( (√5)+(√3))) =∞//
$$\left.\mathrm{3}\right).\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{\mathrm{5x}+\mathrm{2}}\:−\sqrt{\mathrm{3x}−\mathrm{1}}\:\right)×\frac{\left.\sqrt{\mathrm{5x}+\mathrm{2}}\:+\sqrt{\mathrm{3x}−\mathrm{1}}\:\right)}{\left.\:\sqrt{\mathrm{5x}+\mathrm{2}}\:×\sqrt{\mathrm{3x}−\mathrm{1}}\:\right)} \\ $$$$\:\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\left(\mathrm{5x}+\mathrm{2}\right)−\left(\mathrm{3x}−\mathrm{1}\right)}{\left.\:\sqrt{\mathrm{5x}+\mathrm{2}}\:+\sqrt{\mathrm{3x}−\mathrm{1}}\:\right)}\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{2x}+\mathrm{3}}{\left.\:\sqrt{\mathrm{5x}+\mathrm{2}}\:+\sqrt{\mathrm{3x}−\mathrm{1}}\:\right)} \\ $$$$\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{2x}/\sqrt{\mathrm{x}}+\mathrm{3}/\sqrt{\mathrm{x}}}{\left.\:\sqrt{\mathrm{5x}/\mathrm{x}+\mathrm{2}/\mathrm{x}}\:+\sqrt{\mathrm{3x}/\mathrm{x}−\mathrm{1}/\mathrm{x}}\:\right)} \\ $$$$\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{2}\sqrt{\mathrm{x}}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{x}}}}{\left.\:\sqrt{\mathrm{5}+\frac{\mathrm{2}}{\mathrm{x}}}\:+\sqrt{\mathrm{3}−\frac{\mathrm{1}}{\mathrm{x}}}\:\right)} \\ $$$$=\:\frac{\mathrm{2}\sqrt{\infty}+\frac{\mathrm{3}}{\:\sqrt{\infty}}}{\left.\:\sqrt{\mathrm{5}+\frac{\mathrm{2}}{\infty}}\:+\sqrt{\mathrm{3}−\frac{\mathrm{1}}{\infty}}\:\right)}\:=\frac{\infty+\mathrm{0}}{\:\sqrt{\mathrm{5}+\mathrm{0}}\:+\sqrt{\mathrm{3}−\mathrm{0}}} \\ $$$$\:=\frac{\infty}{\:\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}}\:=\infty// \\ $$
Answered by DELETED last updated on 13/Aug/21
4). lim_(x→∞) ((√(4x^2 +x+3)) −(√(3x^2 +2x+5))=...? = lim_(x→∞) ((√(4x^2 +x+3)) −(√(3x^2 +2x+5 ))×(((√(4x^2 +x+3)) +(√(3x^2 +2x+5)))/( (√(4x^2 +x+3)) +(√(3x^2 +2x+5)))) =lim_(x→∞) (((4x^2 +x+3)−(3x^2 +2x+5))/( (√(4x^2 +x+3)) +(√(3x^2 +2x+5 )))) =lim_(x→∞) ((x^2 −x−2)/( (√(4x^2 +x+3)) +(√(3x^2 +2x+5 )))) =lim_(x→∞) ((x^2 /x−x/x−2/x)/( (√(4x^2 /x^2 +x/x^2 +3/x^2 )) +(√(3x^2 /x^2 +2x/x^2 +5/x^2 )))) =lim_(x→∞) ((x−1−2/x)/( (√(4+1/x+3/x^2 +(√(3+2/x+5/x^2 )))))) = ((∞−1−2/∞)/( (√(4+1/∞+3/∞^2 +(√(3+2/∞+5/∞^2 )))))) =((∞−1−0)/( (√(4+0+0))+(√(3+0+0)))) =(∞/(2+(√3))) =∞//
$$\left.\mathrm{4}\right).\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{x}+\mathrm{3}}\:−\sqrt{\mathrm{3x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{5}}=…?\right. \\ $$$$\:\:\:=\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{x}+\mathrm{3}}\:−\sqrt{\mathrm{3x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{5}\:}×\frac{\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{x}+\mathrm{3}}\:+\sqrt{\mathrm{3x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{5}}}{\:\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{x}+\mathrm{3}}\:+\sqrt{\mathrm{3x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{5}}}\right. \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{4x}^{\mathrm{2}} +\mathrm{x}+\mathrm{3}\right)−\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{5}\right)}{\:\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{x}+\mathrm{3}}\:+\sqrt{\mathrm{3x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{5}\:}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{2}}{\:\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{x}+\mathrm{3}}\:+\sqrt{\mathrm{3x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{5}\:}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} /\mathrm{x}−\mathrm{x}/\mathrm{x}−\mathrm{2}/\mathrm{x}}{\:\sqrt{\mathrm{4x}^{\mathrm{2}} /\mathrm{x}^{\mathrm{2}} +\mathrm{x}/\mathrm{x}^{\mathrm{2}} +\mathrm{3}/\mathrm{x}^{\mathrm{2}} }\:+\sqrt{\mathrm{3x}^{\mathrm{2}} /\mathrm{x}^{\mathrm{2}} +\mathrm{2x}/\mathrm{x}^{\mathrm{2}} +\mathrm{5}/\mathrm{x}^{\mathrm{2}} \:}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}−\mathrm{1}−\mathrm{2}/\mathrm{x}}{\:\sqrt{\mathrm{4}+\mathrm{1}/\mathrm{x}+\mathrm{3}/\mathrm{x}^{\mathrm{2}} +\sqrt{\mathrm{3}+\mathrm{2}/\mathrm{x}+\mathrm{5}/\mathrm{x}^{\mathrm{2}} }}} \\ $$$$=\:\frac{\infty−\mathrm{1}−\mathrm{2}/\infty}{\:\sqrt{\mathrm{4}+\mathrm{1}/\infty+\mathrm{3}/\infty^{\mathrm{2}} +\sqrt{\mathrm{3}+\mathrm{2}/\infty+\mathrm{5}/\infty^{\mathrm{2}} }}} \\ $$$$=\frac{\infty−\mathrm{1}−\mathrm{0}}{\:\sqrt{\mathrm{4}+\mathrm{0}+\mathrm{0}}+\sqrt{\mathrm{3}+\mathrm{0}+\mathrm{0}}}\:=\frac{\infty}{\mathrm{2}+\sqrt{\mathrm{3}}}\:=\infty// \\ $$$$ \\ $$

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