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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-




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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