lim-x-1-x-x-1-x-5-32-x-1-

Question Number 223990 by RoseAli last updated on 12/Aug/25

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{5}} −\mathrm{32}\:}{{x}−\mathrm{1}} \\ $$

Commented by prathita last updated on 12/Aug/25

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}} \\ $$

Answered by vnm last updated on 12/Aug/25

x=1+t, t→0 (((1+t)(1+t+(1/(1+t)))^5 −32)/(1+t−1))=(((1+t)(1+t+1−t+o(t))^5 −32)/t)= (((1+t)(2+o(t))^5 −32)/t)=(((1+t)(32+o(t))−32)/t)=((o(t))/t)+32+o(t)→^(t→0) 32

$${x}=\mathrm{1}+{t},\:\:{t}\rightarrow\mathrm{0} \\ $$$$\frac{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right)^{\mathrm{5}} −\mathrm{32}}{\mathrm{1}+{t}−\mathrm{1}}=\frac{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}+\mathrm{1}−{t}+{o}\left({t}\right)\right)^{\mathrm{5}} −\mathrm{32}}{{t}}= \\ $$$$\frac{\left(\mathrm{1}+{t}\right)\left(\mathrm{2}+{o}\left({t}\right)\right)^{\mathrm{5}} −\mathrm{32}}{{t}}=\frac{\left(\mathrm{1}+{t}\right)\left(\mathrm{32}+{o}\left({t}\right)\right)−\mathrm{32}}{{t}}=\frac{{o}\left({t}\right)}{{t}}+\mathrm{32}+{o}\left({t}\right)\overset{{t}\rightarrow\mathrm{0}} {\rightarrow}\mathrm{32} \\ $$

Answered by MirHasibulHossain last updated on 12/Aug/25

=lim_(x→1) (((x+(1/x))^5 −((32)/x))/(1−(1/x))) [∵ Setting: (((1/x)×(Numerator))/((1/x)×(Denominator))) ] =lim_(x→1) ((5(x+(1/x))^4 (1−(1/x^2 ))+((32)/x^2 ))/(0+(1/x^2 ))) [∵ L′Ho^“ pital′s Rule] =((5×2^4 ×0+32)/1) =32 (answer)

$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{5}} −\frac{\mathrm{32}}{\mathrm{x}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}}\:\:\:\:\:\left[\because\:\mathrm{Setting}:\:\:\:\frac{\frac{\mathrm{1}}{\mathrm{x}}×\left(\mathrm{Numerator}\right)}{\frac{\mathrm{1}}{\mathrm{x}}×\left(\mathrm{Denominator}\right)}\:\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{5}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{4}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)+\frac{\mathrm{32}}{\mathrm{x}^{\mathrm{2}} }}{\mathrm{0}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}\:\:\:\:\:\:\:\:\left[\because\:\mathrm{L}'\mathrm{H}\overset{“} {\mathrm{o}pital}'\mathrm{s}\:\mathrm{Rule}\right] \\ $$$$=\frac{\mathrm{5}×\mathrm{2}^{\mathrm{4}} ×\mathrm{0}+\mathrm{32}}{\mathrm{1}} \\ $$$$=\mathrm{32}\:\:\left(\mathrm{answer}\right) \\ $$

Answered by mehdee7396 last updated on 13/Aug/25

hop⇒lim_(x→1) (((x+(1/x))^5 +5x(1−(1/x^2 ))(x+(1/x))^4 )/1)=32 ✓

$${hop}\Rightarrow{li}\underset{{x}\rightarrow\mathrm{1}} {{m}}\frac{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{5}} +\mathrm{5}{x}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{4}} }{\mathrm{1}}=\mathrm{32}\:\checkmark \\ $$$$ \\ $$

Answered by grigoriy last updated on 13/Aug/25

lim_(x→1) ((x(x+(1/x))^5 −32 )/(x−1)) lim_(x→1) ((x(x + (1/x)) − 32)/(x − 1)) = {((1∙(1 + (1/1)) − 32)/(1 − 1))} = {((−30)/0)} = ∞; lim_(x→1) ((x(x + (1/x)) − 32)/(x − 1)) = ∞;

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{5}} −\mathrm{32}\:}{{x}−\mathrm{1}} \\ $$$$ \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}\left({x}\:+\:\frac{\mathrm{1}}{{x}}\right)\:−\:\mathrm{32}}{{x}\:−\:\mathrm{1}}\:=\:\left\{\frac{\mathrm{1}\centerdot\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{1}}\right)\:−\:\mathrm{32}}{\mathrm{1}\:−\:\mathrm{1}}\right\}\:=\:\left\{\frac{−\mathrm{30}}{\mathrm{0}}\right\}\:=\:\infty; \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}\left({x}\:+\:\frac{\mathrm{1}}{{x}}\right)\:−\:\mathrm{32}}{{x}\:−\:\mathrm{1}}\:=\:\infty; \\ $$

Commented by AgniMath last updated on 14/Aug/25

$${Are}\:{you}\:{serious}? \\ $$

Answered by profcedricjunior last updated on 15/Aug/25

lim_(x→1) ((x(x+(1/x))^5 −32)/(x−1))=((2^5 −2^5 )/(1−1))=FI to apply hospital theorem we have =>lim_(x→1) {(x+(1/x))^5 +5x(1−(1/x^2 ))(x+(1/x))^4 }=2^5 =32

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\boldsymbol{{x}}\left(\boldsymbol{{x}}+\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)^{\mathrm{5}} −\mathrm{32}}{\boldsymbol{{x}}−\mathrm{1}}=\frac{\mathrm{2}^{\mathrm{5}} −\mathrm{2}^{\mathrm{5}} }{\mathrm{1}−\mathrm{1}}=\boldsymbol{{FI}} \\ $$$$\boldsymbol{{to}}\:\boldsymbol{{apply}}\:\boldsymbol{{hospital}}\:\boldsymbol{{theorem}}\:\boldsymbol{{we}}\:\boldsymbol{{have}} \\ $$$$=>\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left\{\left(\boldsymbol{{x}}+\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)^{\mathrm{5}} +\mathrm{5}\boldsymbol{{x}}\left(\mathrm{1}−\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} }\right)\left(\boldsymbol{{x}}+\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)^{\mathrm{4}} \right\}=\mathrm{2}^{\mathrm{5}} =\mathrm{32}\: \\ $$

Facebook Tweet Pin

lim-x-1-x-x-1-x-5-32-x-1-

Leave a Reply Cancel reply