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Question Number 75931 by Rio Michael last updated on 21/Dec/19

Evaluate a. lim_(x→∞) (((x^2 +3)/(27x^2 −1)))^(1/3) b. lim_(x→−∞) ((x−2)/(√(x^2 +1))) c. lim_(x→∞) ((x^2 +2)/(2x−3)) d. lim_(x→∞) 2x + 1 − (√(4x^2 +5))

$${Evaluate} \\ $$$${a}.\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\frac{{x}^{\mathrm{2}} +\mathrm{3}}{\mathrm{27}{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${b}.\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{{x}−\mathrm{2}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${c}.\underset{{x}\rightarrow\infty} {\:\mathrm{lim}}\frac{{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}−\mathrm{3}} \\ $$$${d}.\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{2}{x}\:+\:\mathrm{1}\:−\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}} \\ $$

Commented by turbo msup by abdo last updated on 21/Dec/19

a)we have lim_(x→∞) ((x^2 +3)/(27x^2 −1)) =(1/(27)) ⇒lim_(x→∞) ^3 (√((x^2 +3)/(27x^2 −1))) =(1/((^3 (√(27))))) =(1/3) b)lim_(x→−∞) ((x−2)/(√(x^2 +1))) =lim_(x→−∞) ((x−2)/((√(x^2 (1+(1/x^2 )))))) =lim_(x→−∞) ((x−2)/(∣x∣(√(1+(1/x^2 ))))) =lim_(x→−∞) (x/(−x))=−1

$$\left.{a}\right){we}\:{have}\:{lim}_{{x}\rightarrow\infty} \:\:\frac{{x}^{\mathrm{2}} +\mathrm{3}}{\mathrm{27}{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{27}}\:\Rightarrow{lim}_{{x}\rightarrow\infty} \overset{\mathrm{3}} {\:}\sqrt{\frac{{x}^{\mathrm{2}} +\mathrm{3}}{\mathrm{27}{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\frac{\mathrm{1}}{\left(^{\mathrm{3}} \sqrt{\mathrm{27}}\right)}\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\left.{b}\right){lim}_{{x}\rightarrow−\infty} \:\:\:\frac{{x}−\mathrm{2}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$={lim}_{{x}\rightarrow−\infty} \:\:\frac{{x}−\mathrm{2}}{\left.\sqrt{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right.}\right)} \\ $$$$={lim}_{{x}\rightarrow−\infty} \:\:\:\:\:\frac{{x}−\mathrm{2}}{\mid{x}\mid\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}} \\ $$$$={lim}_{{x}\rightarrow−\infty} \:\:\:\frac{{x}}{−{x}}=−\mathrm{1} \\ $$

Commented by Rio Michael last updated on 21/Dec/19

thanks sir

$${thanks}\:{sir} \\ $$

Commented by turbo msup by abdo last updated on 21/Dec/19

c)lim_(x→∞) ((x^2 +2)/(2x−3)) =lim_(x→∞) (x^2 /(2x)) =lim_(x→∞) (x/2) =∞ let f(x)=2x+1−(√(4x^2 +5)) ⇒f(x)=2x+1−∣2x∣(√(1+(5/(4x^2 )))) ∼2x+1−∣2x∣(1+(5/(8x^2 ))) f(x)∼ 2x+1−2∣x∣−((10)/(8∣x∣)) so f(x)∼1−((10)/(8x)) if (x→+∞) and f(x)∼4x+1 +((10)/(8x)) (x→−∞) ⇒ lim_(x→+∞) f(x)=1 and lim_(x→−∞) f(x)=−∞

$$\left.{c}\right){lim}_{{x}\rightarrow\infty} \:\:\frac{{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}−\mathrm{3}}\:={lim}_{{x}\rightarrow\infty} \:\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}} \\ $$$$={lim}_{{x}\rightarrow\infty} \:\:\frac{{x}}{\mathrm{2}}\:=\infty \\ $$$${let}\:{f}\left({x}\right)=\mathrm{2}{x}+\mathrm{1}−\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{2}{x}+\mathrm{1}−\mid\mathrm{2}{x}\mid\sqrt{\mathrm{1}+\frac{\mathrm{5}}{\mathrm{4}{x}^{\mathrm{2}} }} \\ $$$$\sim\mathrm{2}{x}+\mathrm{1}−\mid\mathrm{2}{x}\mid\left(\mathrm{1}+\frac{\mathrm{5}}{\mathrm{8}{x}^{\mathrm{2}} }\right) \\ $$$${f}\left({x}\right)\sim\:\mathrm{2}{x}+\mathrm{1}−\mathrm{2}\mid{x}\mid−\frac{\mathrm{10}}{\mathrm{8}\mid{x}\mid}\:{so} \\ $$$${f}\left({x}\right)\sim\mathrm{1}−\frac{\mathrm{10}}{\mathrm{8}{x}}\:\:{if}\:\left({x}\rightarrow+\infty\right) \\ $$$${and}\:{f}\left({x}\right)\sim\mathrm{4}{x}+\mathrm{1}\:+\frac{\mathrm{10}}{\mathrm{8}{x}}\:\left({x}\rightarrow−\infty\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=\mathrm{1}\:{and} \\ $$$${lim}_{{x}\rightarrow−\infty} \:{f}\left({x}\right)=−\infty \\ $$

Commented by turbo msup by abdo last updated on 21/Dec/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

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