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Question Number 134822 by abdurehime last updated on 07/Mar/21

Answered by greg_ed last updated on 07/Mar/21

p=lim_(x→1) ((x^4 −(√x))/( (√x)−1)) by L′Hopital′s rule p=lim_(x→1) ((4x^3 −(1/(2(√x))))/(1/(2 (√x)))) plug in x=1 lim_(x→1) ((x^4 −(√x))/( (√x)−1))=7

$$\mathrm{p}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{4}} −\sqrt{{x}}}{\:\sqrt{{x}}−\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{L}}'\boldsymbol{\mathrm{Hopital}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{rule}} \\ $$$$\mathrm{p}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{4}{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}}{\frac{\mathrm{1}}{\mathrm{2}\:\sqrt{{x}}}}\: \\ $$$$\boldsymbol{\mathrm{plug}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{{x}}=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{4}} −\sqrt{{x}}}{\:\sqrt{{x}}−\mathrm{1}}=\mathrm{7} \\ $$

Answered by EDWIN88 last updated on 07/Mar/21

lim_(x→1) ((x^4 −(√x))/( (√x)−1)) let (√x) = u then lim_(u→1) ((u^8 −u)/(u−1)) = lim_(u→1) ((u(u^7 −1))/(u−1)) = lim_(u→1) u(u^6 +u^5 +u^4 +u^3 +u^2 +u+1) = 1×(1+1+1+...+1)_(7 times) = 7×1 = 7

$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{4}} −\sqrt{\mathrm{x}}}{\:\sqrt{\mathrm{x}}−\mathrm{1}}\: \\ $$$$\mathrm{let}\:\sqrt{\mathrm{x}}\:=\:\mathrm{u}\:\mathrm{then}\:\underset{\mathrm{u}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{u}^{\mathrm{8}} −\mathrm{u}}{\mathrm{u}−\mathrm{1}}\:=\:\underset{\mathrm{u}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{u}\left(\mathrm{u}^{\mathrm{7}} −\mathrm{1}\right)}{\mathrm{u}−\mathrm{1}} \\ $$$$=\:\underset{\mathrm{u}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\mathrm{u}\left(\mathrm{u}^{\mathrm{6}} +\mathrm{u}^{\mathrm{5}} +\mathrm{u}^{\mathrm{4}} +\mathrm{u}^{\mathrm{3}} +\mathrm{u}^{\mathrm{2}} +\mathrm{u}+\mathrm{1}\right) \\ $$$$=\:\mathrm{1}×\underset{\mathrm{7}\:\mathrm{times}} {\underbrace{\left(\mathrm{1}+\mathrm{1}+\mathrm{1}+...+\mathrm{1}\right)}}\:=\:\mathrm{7}×\mathrm{1}\:=\:\mathrm{7} \\ $$

Answered by EDWIN88 last updated on 07/Mar/21

lim_(x→∞) ((√(9x^4 +1))/(x^2 −3x+5)) = lim_(x→∞) ((x^2 (√(9+(1/x^4 ))))/(x^2 (1−(3/x)+(5/x^2 )))) = lim_(x→∞) ((√(9+(1/x^2 )))/(1−(3/x)+(5/x^2 ))) ; let (1/x) = t and t→0 = lim_(t→0) ((√(9+t^2 ))/(1−3t+5t^2 )) = (3/1)= 3

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{9x}^{\mathrm{4}} +\mathrm{1}}}{\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{5}}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} \:\sqrt{\mathrm{9}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} }\right)} \\ $$$$=\:\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{9}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} }}\:;\:\mathrm{let}\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{t}\:\mathrm{and}\:\mathrm{t}\rightarrow\mathrm{0} \\ $$$$=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{9}+\mathrm{t}^{\mathrm{2}} }}{\mathrm{1}−\mathrm{3t}+\mathrm{5t}^{\mathrm{2}} }\:=\:\frac{\mathrm{3}}{\mathrm{1}}=\:\mathrm{3} \\ $$

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