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Question Number 132483 by abdullahquwatan last updated on 14/Feb/21
lim_(x→−2) (((√(x^2 +2x+4))−x^2 +2)/(x^5 +32)) no hospital
$$\underset{{x}\rightarrow−\mathrm{2}} {\mathrm{lim}}\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}−{x}^{\mathrm{2}} +\mathrm{2}}{{x}^{\mathrm{5}} +\mathrm{32}}\:\mathrm{no}\:\mathrm{hospital} \\ $$
Commented by EDWIN88 last updated on 14/Feb/21
what hospital?
$$\mathrm{what}\:\mathrm{hospital}? \\ $$
Commented by abdullahquwatan last updated on 14/Feb/21
l′hopital theorem
$$\mathrm{l}'\mathrm{hopital}\:\mathrm{theorem} \\ $$
Answered by EDWIN88 last updated on 14/Feb/21
let x+2 = t ∧t→0 lim_(t→0) (((√((t−2)^2 +2(t−2)+4)) +2−(t−2)^2 )/((t−2)^5 +32)) lim_(t→0) (((√(t^2 −2t+4)) −(t^2 −4t+2))/((t−2)^5 +32))= lim_(t→0) (((t^2 −2t+4)−(t^2 −4t+2)^2 )/((t−2)^5 +32)) × lim_(t→0) (1/( (√(t^2 −2t+4))+t^2 −4t+2)) = (1/4)×lim_(t→0) ((t^2 −2t+4−(t^4 +4t^2 +4−8t^3 +4t^2 −16t))/(t^5 −10t^4 +20t^3 −80t^2 +80t)) =(1/4)×lim_(t→0) ((−t^4 +8t^3 −7t^2 +14t)/(t^5 −10t^4 +20t^3 −80t^2 +80t)) = (1/4)×((14)/(80)) = (7/(160))
$$\mathrm{let}\:\mathrm{x}+\mathrm{2}\:=\:\mathrm{t}\:\wedge\mathrm{t}\rightarrow\mathrm{0} \\ $$$$\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\left(\mathrm{t}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{t}−\mathrm{2}\right)+\mathrm{4}}\:+\mathrm{2}−\left(\mathrm{t}−\mathrm{2}\right)^{\mathrm{2}} }{\left(\mathrm{t}−\mathrm{2}\right)^{\mathrm{5}} +\mathrm{32}} \\ $$$$\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{4}}\:−\left(\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{2}\right)}{\left(\mathrm{t}−\mathrm{2}\right)^{\mathrm{5}} +\mathrm{32}}= \\ $$$$\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{4}\right)−\left(\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{2}\right)^{\mathrm{2}} }{\left(\mathrm{t}−\mathrm{2}\right)^{\mathrm{5}} +\mathrm{32}}\:×\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{4}}+\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{2}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}×\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{4}−\left(\mathrm{t}^{\mathrm{4}} +\mathrm{4t}^{\mathrm{2}} +\mathrm{4}−\mathrm{8t}^{\mathrm{3}} +\mathrm{4t}^{\mathrm{2}} −\mathrm{16t}\right)}{\mathrm{t}^{\mathrm{5}} −\mathrm{10t}^{\mathrm{4}} +\mathrm{20t}^{\mathrm{3}} −\mathrm{80t}^{\mathrm{2}} +\mathrm{80t}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{t}^{\mathrm{4}} +\mathrm{8t}^{\mathrm{3}} −\mathrm{7t}^{\mathrm{2}} +\mathrm{14t}}{\mathrm{t}^{\mathrm{5}} −\mathrm{10t}^{\mathrm{4}} +\mathrm{20t}^{\mathrm{3}} −\mathrm{80t}^{\mathrm{2}} +\mathrm{80t}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{14}}{\mathrm{80}}\:=\:\frac{\mathrm{7}}{\mathrm{160}} \\ $$
Commented by abdullahquwatan last updated on 14/Feb/21
thank you so much
$${thank}\:{you}\:{so}\:{much} \\ $$
Answered by bemath last updated on 14/Feb/21
If by L′Ho^ pital rule lim_(x→−2) ((((x+1)/( (√(x^2 +2x+4)))) −2x)/(5x^4 )) = ((((−1)/( (√4)))+4)/(80)) = ((−(1/2)+4)/(80))=((−1+8)/(160))=(7/(160))
$$\mathrm{If}\:\mathrm{by}\:\mathrm{L}'\mathrm{H}\ddot {\mathrm{o}pital}\:\mathrm{rule} \\ $$$$\:\underset{{x}\rightarrow−\mathrm{2}} {\mathrm{lim}}\:\frac{\frac{\mathrm{x}+\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{4}}}\:−\mathrm{2x}}{\mathrm{5x}^{\mathrm{4}} }\:= \\ $$$$\:\frac{\frac{−\mathrm{1}}{\:\sqrt{\mathrm{4}}}+\mathrm{4}}{\mathrm{80}}\:=\:\frac{−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{4}}{\mathrm{80}}=\frac{−\mathrm{1}+\mathrm{8}}{\mathrm{160}}=\frac{\mathrm{7}}{\mathrm{160}} \\ $$

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