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Question Number 199451 by hardmath last updated on 03/Nov/23

Find: 1. lim_(n→∞) (((5n − 25)/(3n + 15)))^(1/(5n)) 2. lim_(x→∞) (((x^3 + 3x^2 + 1))^(1/3) − ((x^3 − 3x^2 + 1))^(1/3) ) 3. lim_(x→0) ((cos 4x^3 − 1)/(sin^6 2x))

$$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{5}\boldsymbol{\mathrm{n}}}]{\frac{\mathrm{5n}\:−\:\mathrm{25}}{\mathrm{3n}\:+\:\mathrm{15}}} \\ $$$$\mathrm{2}.\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{3x}^{\mathrm{2}} \:+\:\mathrm{1}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{3x}^{\mathrm{2}} \:+\:\mathrm{1}}\:\right) \\ $$$$\mathrm{3}.\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{4x}^{\mathrm{3}} \:−\:\mathrm{1}}{\mathrm{sin}^{\mathrm{6}} \:\mathrm{2x}} \\ $$

Answered by cortano12 last updated on 04/Nov/23

(2) lim_(x→∞) ((6x^2 )/( (((x^3 +3x^2 +1)^2 ))^(1/3) +(((x^3 +3x^2 +1)(x^3 −3x^2 +1)))^(1/3) +(((x^3 −3x^2 +1)^2 ))^(1/3) )) = lim_(x→∞) ((6x^2 )/(x^2 [ (((1+(3/x)+(1/x^3 ))^2 ))^(1/3) +(((1−(3/x)+(1/x^3 ))(1+(3/x)+(1/x^3 ))))^(1/3) +(((1−(3/x)+(1/x^3 ))^2 ))^(1/3) ])) = (6/3)=2

$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{6x}^{\mathrm{2}} }{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{\left(\mathrm{x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{1}\right)}+\sqrt[{\mathrm{3}}]{\left(\mathrm{x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$$\:\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{6x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} \:\left[\:\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)}+\sqrt[{\mathrm{3}}]{\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)^{\mathrm{2}} \:}\right]} \\ $$$$\:=\:\frac{\mathrm{6}}{\mathrm{3}}=\mathrm{2} \\ $$$$ \\ $$

Commented by hardmath last updated on 04/Nov/23

thank you ser

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

Answered by cortano12 last updated on 04/Nov/23

(3) lim_(x→0) ((cos 4x^3 −1)/(sin^6 2x)) = lim_(x→0) ((−sin^2 (4x^3 ))/((cos 4x^3 +1)sin^6 (2x))) = −(1/2) lim_(x→0) (((4x^3 )^2 )/((2x)^6 )) = −(1/2).((16)/(64)) = −(1/8)

$$\left(\mathrm{3}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{4x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{sin}\:^{\mathrm{6}} \mathrm{2x}} \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{4x}^{\mathrm{3}} \right)}{\left(\mathrm{cos}\:\mathrm{4x}^{\mathrm{3}} +\mathrm{1}\right)\mathrm{sin}\:^{\mathrm{6}} \left(\mathrm{2x}\right)} \\ $$$$\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{4x}^{\mathrm{3}} \right)^{\mathrm{2}} }{\left(\mathrm{2x}\right)^{\mathrm{6}} }\:=\:−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{16}}{\mathrm{64}} \\ $$$$\:\:=\:−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$ \\ $$

Commented by hardmath last updated on 04/Nov/23

thank you ser

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

Answered by cortano12 last updated on 04/Nov/23

(1) lim_(n→∞) (((5n−25)/(3n+15)))^(1/(5n)) = e^(lim_(n→∞) (((5n−25)/(3n+15))−1).(1/(5n))) = e^(lim_(n→∞) (((2n−40)/(15n(n+5))))) = e^0 =1

$$\left(\mathrm{1}\right)\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{5n}−\mathrm{25}}{\mathrm{3n}+\mathrm{15}}\right)^{\frac{\mathrm{1}}{\mathrm{5n}}} \\ $$$$\:\:=\:\mathrm{e}\:^{\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{5n}−\mathrm{25}}{\mathrm{3n}+\mathrm{15}}−\mathrm{1}\right).\frac{\mathrm{1}}{\mathrm{5n}}} \\ $$$$\:\:=\:\mathrm{e}\:^{\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2n}−\mathrm{40}}{\mathrm{15n}\left(\mathrm{n}+\mathrm{5}\right)}\right)} =\:\mathrm{e}^{\mathrm{0}} =\mathrm{1} \\ $$

Commented by hardmath last updated on 04/Nov/23

thank you ser

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

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