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Question Number 183350 by Spillover last updated on 25/Dec/22

Find (a)lim_(x→∞) ((3x+2)/(x^2 −x+1)) (b)lim_(x→∞) ((√(x^2 −1))/(2x+1)) (c)lim_(x→5) (((√(3x+1)) −4)/(x−5))

$${Find}\: \\ $$$$\left({a}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\:\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$\left({b}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$$\left({c}\right)\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{3}{x}+\mathrm{1}}\:−\mathrm{4}}{{x}−\mathrm{5}} \\ $$

Commented by CElcedricjunior last updated on 25/Dec/22

a)0 b)(1/2) c)(3/8)

$$\left.\boldsymbol{{a}}\right)\mathrm{0} \\ $$$$\left.\boldsymbol{{b}}\right)\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left.\boldsymbol{{c}}\right)\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$ \\ $$

Commented by Spillover last updated on 25/Dec/22

thank you

$${thank}\:{you} \\ $$

Answered by Spillover last updated on 25/Dec/22

(a)lim_(x→∞) ((3x+2)/(x^2 −x+1)) Divide by x^2 up and down lim_(x→∞) (((3/x)+(2/x))/(1−(1/x)+(1/x^2 ))) x→∞ x=0 ((0+0)/(1−0+0)) = 0

$$ \\ $$$$\left({a}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:\: \\ $$$${Divide}\:{by}\:{x}^{\mathrm{2}} \:{up}\:{and}\:{down} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\frac{\mathrm{3}}{{x}}+\frac{\mathrm{2}}{{x}}}{\mathrm{1}−\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\:\:\:\:\:{x}\rightarrow\infty\:\:\:{x}=\mathrm{0} \\ $$$$\frac{\mathrm{0}+\mathrm{0}}{\mathrm{1}−\mathrm{0}+\mathrm{0}}\:=\:\mathrm{0} \\ $$

Answered by Spillover last updated on 25/Dec/22

(b)lim_(x→∞) ((√(x^2 −1))/(2x+1)) divide by x up and down lim_(x→∞) ((√(x^2 −1))/(2x+1))= lim_(x→∞) (((√(x^2 −1))/x)/((2x+1)/x)) lim_(x→∞) ((√(1−(1/x^2 )))/(2+(1/x))) x→∞ ((√(1−0))/(2+0)) =(1/2)

$$\left({b}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$${divide}\:{by}\:{x}\:\:{up}\:{and}\:{down} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}{x}+\mathrm{1}}=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}}}{\frac{\mathrm{2}{x}+\mathrm{1}}{{x}}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}{\mathrm{2}+\frac{\mathrm{1}}{{x}}}\:\:\:\: \\ $$$${x}\rightarrow\infty\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{1}−\mathrm{0}}}{\mathrm{2}+\mathrm{0}}\:\:\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

Answered by Spillover last updated on 25/Dec/22

(c)lim_(x→5) (((√(3x+1)) −4)/(x−5)) Rationalize numerator lim_(x→5) (((√(3x+1)) −4)/(x−5))×(((√(3x+1)) +4)/( (√(3x+1)) +4)) lim_(x→5) ((3x+1−16)/((x−5)((√(3x+1)) +4))) lim_(x→5) (3/( (√(3x+1)) +4)) x→5 x=5 (3/( (√(3×5+1)) +4)) =(3/8)

$$\left({c}\right)\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{3}{x}+\mathrm{1}}\:−\mathrm{4}}{{x}−\mathrm{5}} \\ $$$${Rationalize}\:{numerator}\: \\ $$$$\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{3}{x}+\mathrm{1}}\:−\mathrm{4}}{{x}−\mathrm{5}}×\frac{\sqrt{\mathrm{3}{x}+\mathrm{1}}\:+\mathrm{4}}{\:\sqrt{\mathrm{3}{x}+\mathrm{1}}\:+\mathrm{4}} \\ $$$$\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\:\:\:\frac{\mathrm{3}{x}+\mathrm{1}−\mathrm{16}}{\left({x}−\mathrm{5}\right)\left(\sqrt{\mathrm{3}{x}+\mathrm{1}}\:+\mathrm{4}\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}{x}+\mathrm{1}}\:+\mathrm{4}}\:\:\:\: \\ $$$${x}\rightarrow\mathrm{5}\:\:\:\:{x}=\mathrm{5} \\ $$$$\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}×\mathrm{5}+\mathrm{1}}\:+\mathrm{4}}\:=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$ \\ $$

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