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Question Number 131882 by Eric002 last updated on 09/Feb/21

lim_(x→0) (x/(x+((x+1))^(1/4) −1))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{{x}+\sqrt[{\mathrm{4}}]{{x}+\mathrm{1}}−\mathrm{1}} \\ $$

Answered by liberty last updated on 09/Feb/21

L′Ho^ pital L= lim_(x→0) [ (1/(1+(1/(4 (((x+1)^3 ))^(1/4) )))) ]=(1/(1+(1/4)))= (4/5) or lim_(x→0) [ (x/(x+1+(x/4)−1))] = lim_(x→0) ((4x)/(5x))=(4/5)

$$\:\mathrm{L}'\mathrm{H}\ddot {\mathrm{o}pital}\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}\:\sqrt[{\mathrm{4}}]{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} }}}\:\right]=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}}=\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{or}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\:\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}+\frac{\mathrm{x}}{\mathrm{4}}−\mathrm{1}}\right]\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4x}}{\mathrm{5x}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$

Commented by Eric002 last updated on 09/Feb/21

how did you get (x/(x+1+(x/4)−1))

$${how}\:{did}\:{you}\:{get}\:\frac{{x}}{{x}+\mathrm{1}+\frac{{x}}{\mathrm{4}}−\mathrm{1}} \\ $$

Commented by EDWIN88 last updated on 09/Feb/21

Bernoulli equation lim_(x→0) (1+f(x))^n ≈ lim_(x→0) (1+n.f(x)) so lim_(x→0) ((1+x))^(1/4) ≈ lim_(x→0) (1+(x/4))

$$\mathrm{Bernoulli}\:\mathrm{equation}\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+\mathrm{f}\left(\mathrm{x}\right)\right)^{\mathrm{n}} \:\approx\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{n}.\mathrm{f}\left(\mathrm{x}\right)\right) \\ $$$$\:\mathrm{so}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt[{\mathrm{4}}]{\mathrm{1}+\mathrm{x}}\:\approx\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{4}}\right) \\ $$

Commented by Eric002 last updated on 09/Feb/21

thank you sir

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

Answered by EDWIN88 last updated on 09/Feb/21

the another way let ((x+1))^(1/4) = ℓ where ℓ→1 and x=ℓ^4 −1 L=lim_(ℓ→1) ((ℓ^4 −1)/(ℓ^4 +ℓ−2)) = lim_(ℓ→1) (((ℓ^2 +1)(ℓ+1)(ℓ−1))/((ℓ−1)(ℓ^3 +ℓ^2 +ℓ+2))) = lim_(ℓ→1) (((ℓ^2 +1)(ℓ+1))/(ℓ^3 +ℓ^2 +ℓ+2)) = (4/5)

$$\mathrm{the}\:\mathrm{another}\:\mathrm{way} \\ $$$$\:\mathrm{let}\:\sqrt[{\mathrm{4}}]{\mathrm{x}+\mathrm{1}}\:=\:\ell\:\mathrm{where}\:\ell\rightarrow\mathrm{1}\:\mathrm{and}\:\mathrm{x}=\ell^{\mathrm{4}} −\mathrm{1} \\ $$$$\mathrm{L}=\underset{\ell\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\ell^{\mathrm{4}} −\mathrm{1}}{\ell^{\mathrm{4}} +\ell−\mathrm{2}}\:=\:\underset{\ell\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\left(\ell^{\mathrm{2}} +\mathrm{1}\right)\left(\ell+\mathrm{1}\right)\left(\ell−\mathrm{1}\right)}{\left(\ell−\mathrm{1}\right)\left(\ell^{\mathrm{3}} +\ell^{\mathrm{2}} +\ell+\mathrm{2}\right)}\: \\ $$$$\:=\:\underset{\ell\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\left(\ell^{\mathrm{2}} +\mathrm{1}\right)\left(\ell+\mathrm{1}\right)}{\ell^{\mathrm{3}} +\ell^{\mathrm{2}} +\ell+\mathrm{2}}\:=\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 10/Feb/21

let f(x)=(x/(x+^4 (√(1+x))−1)) ⇒f(x)=(x/(x+(1+x)^(1/4) −1)) ⇒f(x)∼(x/(x+1+(x/4)−1))=(x/((5/4)x))=(4/5) ⇒lim_(x⌣)0) f(x)=(4/5)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{x}+^{\mathrm{4}} \sqrt{\mathrm{1}+\mathrm{x}}−\mathrm{1}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{x}+\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{1}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}+\frac{\mathrm{x}}{\mathrm{4}}−\mathrm{1}}=\frac{\mathrm{x}}{\frac{\mathrm{5}}{\mathrm{4}}\mathrm{x}}=\frac{\mathrm{4}}{\mathrm{5}}\:\Rightarrow\mathrm{lim}_{\left.\mathrm{x}\smile\right)\mathrm{0}} \:\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{4}}{\mathrm{5}} \\ $$

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