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Question Number 120211 by bemath last updated on 30/Oct/20

evaluate lim_(x→∞) f(x) and lim_(x→−∞) f(x) for f(x)=(x/( (√(x^2 +1)))).

$$\:{evaluate}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{f}\left({x}\right)\:{and}\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{f}\left({x}\right) \\ $$$${for}\:{f}\left({x}\right)=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}. \\ $$

Answered by Ar Brandon last updated on 30/Oct/20

f(x)=(x/( ∣x∣(√(1+(1/x^2 )))))=((sgn(x))/( (√(1+(1/x^2 ))))) lim_(x→∞) f(x)=1, lim_(x→−∞) f(x)=−1

$${f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\:\mid\mathrm{x}\mid\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}}=\frac{\mathrm{sgn}\left(\mathrm{x}\right)}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}} \\ $$$$\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}{f}\left(\mathrm{x}\right)=\mathrm{1},\:\underset{\mathrm{x}\rightarrow−\infty} {\mathrm{lim}}{f}\left(\mathrm{x}\right)=−\mathrm{1} \\ $$

Commented by Dwaipayan Shikari last updated on 30/Oct/20

বাংলা শিখছো?

Commented by Ar Brandon last updated on 30/Oct/20

না, আমি কিছু মজা করছি������

Answered by benjo_mathlover last updated on 30/Oct/20

Rewrite the expression for f(x) as follows : f(x)=(x/( (√(x^2 (1+(1/x^2 )))))) f(x) = (x/( (√x^2 ) (√(1+(1/x^2 ))))) =(x/(∣x∣(√(1+(1/x^2 ))))) f(x)= ((sgn x)/( (√(1+(1/x^2 ))))) ; [ where sgn x = (x/(∣x∣))= { ((1, if x>0)),((−1, if x<0)) :} The factor (√(1+(1/x^2 ))) approaches 1 as x→∞ or −∞ therefore { ((lim_(x→∞) f(x)=1)),((lim_(x→−∞) f(x)=−1)) :}

$${Rewrite}\:{the}\:{expression}\:{for} \\ $$$${f}\left({x}\right)\:{as}\:{follows}\::\:{f}\left({x}\right)=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}} \\ $$$${f}\left({x}\right)\:=\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} }\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}\:=\frac{{x}}{\mid{x}\mid\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}} \\ $$$${f}\left({x}\right)=\:\frac{{sgn}\:{x}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}\:;\:\left[\:{where}\:{sgn}\:{x}\:=\:\frac{{x}}{\mid{x}\mid}=\begin{cases}{\mathrm{1},\:{if}\:{x}>\mathrm{0}}\\{−\mathrm{1},\:{if}\:{x}<\mathrm{0}}\end{cases}\right. \\ $$$${The}\:{factor}\:\sqrt{\mathrm{1}+\left(\mathrm{1}/{x}^{\mathrm{2}} \right)}\:{approaches}\:\mathrm{1}\:{as}\:{x}\rightarrow\infty\:{or}\:−\infty \\ $$$${therefore}\:\begin{cases}{\underset{{x}\rightarrow\infty} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{1}}\\{\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{f}\left({x}\right)=−\mathrm{1}}\end{cases} \\ $$

Answered by mathmax by abdo last updated on 30/Oct/20

lim_(x→+∞) f(x) =lim_(x→+∞) (x/(x(√(1+(1/x^2 ))))) =lim_(x→+∞) (1/(√(1+(1/x^2 ))))=1 lim_(x→−∞) f(x) =lim_(x→−∞) (x/(−x(√(1+(1/x^2 ))))) =−lim_(x→−∞) (1/(√(1+(1/x^2 ))))=−1 (look that f(x)=(x/(∣x∣(√(1+(1/x^2 ))))) for x≠0)

$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \:\:\frac{\mathrm{x}}{\mathrm{x}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}}\:=\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \frac{\mathrm{1}}{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}}=\mathrm{1} \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow−\infty} \:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{lim}_{\mathrm{x}\rightarrow−\infty} \:\:\frac{\mathrm{x}}{−\mathrm{x}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}}\:=−\mathrm{lim}_{\mathrm{x}\rightarrow−\infty} \:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}}=−\mathrm{1} \\ $$$$\left(\mathrm{look}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mid\mathrm{x}\mid\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}}\:\mathrm{for}\:\mathrm{x}\neq\mathrm{0}\right) \\ $$

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