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Question Number 184794 by Spillover last updated on 11/Jan/23

Evaluate lim_(x→2) ((x^5 −32)/(x^3 −8))

$${Evaluate}\: \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{5}} −\mathrm{32}}{{x}^{\mathrm{3}} −\mathrm{8}} \\ $$

Commented by MJS_new last updated on 11/Jan/23

((20)/3)

$$\frac{\mathrm{20}}{\mathrm{3}} \\ $$

Commented by Spillover last updated on 11/Jan/23

thank you

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

Answered by Spillover last updated on 12/Jan/23

lim_(x→2) ((x^5 −2^5 )/(x^3 −2^3 ))=lim_(x→2) (((x^5 −2^5 )/(x−2))/((x^3 −2^3 )/(x−2))) from lim_(x→a) ((x^n −a^n )/(x−a)) =nax^(n−1) lim_(x→2) (((x^5 −2^5 )/(x−2))/((x^3 −2^3 )/(x−2)))=lim_(x→2) ((x^5 −2^5 )/(x−2))÷lim_(x→2) ((x^3 −2^3 )/(x−2)) (5×2^4 )÷(3×2^2 )=((20)/3)

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}^{\mathrm{5}} −\mathrm{2}^{\mathrm{5}} }{{x}^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} }=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\frac{{x}^{\mathrm{5}} −\mathrm{2}^{\mathrm{5}} }{{x}−\mathrm{2}}}{\frac{{x}^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} }{{x}−\mathrm{2}}} \\ $$$${from} \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{x}^{{n}} −{a}^{{n}} }{{x}−{a}}\:={nax}^{{n}−\mathrm{1}} \\ $$$$\mathrm{li}\underset{{x}\rightarrow\mathrm{2}} {\mathrm{m}}\frac{\frac{{x}^{\mathrm{5}} −\mathrm{2}^{\mathrm{5}} }{{x}−\mathrm{2}}}{\frac{{x}^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} }{{x}−\mathrm{2}}}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}^{\mathrm{5}} −\mathrm{2}^{\mathrm{5}} }{{x}−\mathrm{2}}\boldsymbol{\div}\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} }{{x}−\mathrm{2}} \\ $$$$\left(\mathrm{5}×\mathrm{2}^{\mathrm{4}} \:\right)\boldsymbol{\div}\left(\mathrm{3}×\mathrm{2}^{\mathrm{2}} \right)=\frac{\mathrm{20}}{\mathrm{3}} \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 12/Jan/23

lim_(x→2) ((x^5 −32)/(x^3 −8)) =lim_(x→2) (((x−2)(x^4 +2x^3 +4x^2 +8x+16))/((x−2)(x^2 +2x+4))) =((lim_(x→2) (x^4 +2x^3 +4x^2 +8x+16))/(lim_(x→2) (x^2 +2x+4))) =((2^4 +2∙2^3 +4∙2^2 +8∙2+16)/(2^2 +2∙2+4)) =((5∙2^4 )/(3∙2^2 ))=(5/3)∙2^(4−2) =((20)/3)

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{5}} −\mathrm{32}}{{x}^{\mathrm{3}} −\mathrm{8}} \\ $$$$=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\cancel{\left({x}−\mathrm{2}\right)}\left({x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{16}\right)}{\cancel{\left({x}−\mathrm{2}\right)}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)} \\ $$$$=\frac{\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left({x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{16}\right)}{\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)} \\ $$$$=\frac{\mathrm{2}^{\mathrm{4}} +\mathrm{2}\centerdot\mathrm{2}^{\mathrm{3}} +\mathrm{4}\centerdot\mathrm{2}^{\mathrm{2}} +\mathrm{8}\centerdot\mathrm{2}+\mathrm{16}}{\mathrm{2}^{\mathrm{2}} +\mathrm{2}\centerdot\mathrm{2}+\mathrm{4}} \\ $$$$=\frac{\mathrm{5}\centerdot\mathrm{2}^{\mathrm{4}} }{\mathrm{3}\centerdot\mathrm{2}^{\mathrm{2}} }=\frac{\mathrm{5}}{\mathrm{3}}\centerdot\mathrm{2}^{\mathrm{4}−\mathrm{2}} =\frac{\mathrm{20}}{\mathrm{3}} \\ $$

Commented by Spillover last updated on 12/Jan/23

nice approach

$${nice}\:{approach} \\ $$

Answered by BaliramKumar last updated on 14/Jan/23

lim_(x→2) ((x^5 −32)/(x^3 −8)) lim_(x→2) ((5x^4 )/(3x^2 )) =((5x^2 )/3) = ((5(2)^2 )/3) = ((20)/3)

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{5}} −\mathrm{32}}{{x}^{\mathrm{3}} −\mathrm{8}} \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\:\:\:\:\frac{\mathrm{5}{x}^{\mathrm{4}} }{\mathrm{3}{x}^{\mathrm{2}} }\:\:\:=\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{3}}\:=\:\:\frac{\mathrm{5}\left(\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{3}}\:=\:\frac{\mathrm{20}}{\mathrm{3}} \\ $$$$ \\ $$

Answered by ajfour last updated on 14/Jan/23

lim_(x→a) ((x^n −a^n )/(x−a))=na^(n−1) hence =((5(2^4 ))/(3(2^2 )))=((20)/3)

$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{x}^{{n}} −{a}^{{n}} }{{x}−{a}}={na}^{{n}−\mathrm{1}} \\ $$$${hence}\:\:\:=\frac{\mathrm{5}\left(\mathrm{2}^{\mathrm{4}} \right)}{\mathrm{3}\left(\mathrm{2}^{\mathrm{2}} \right)}=\frac{\mathrm{20}}{\mathrm{3}} \\ $$

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