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1-lim-x-0-5-x-1-3-3-x-1-4-x-1-6-4-x-1-3-x-1-24-2-lim-x-5-x-1-3-3-x-1-4-x-1-6-4-x-1-3-x-1-24-




Question Number 125056 by benjo_mathlover last updated on 08/Dec/20
 (1) lim_(x→0)  ((5 (x)^(1/3)  −3 (x)^(1/4)  −(x)^(1/6)  )/(4 (x)^(1/3)  −(x)^(1/(24)) )) ?  (2) lim_(x→∞)  ((5 (x)^(1/3)  −3 (x)^(1/4)  −(x)^(1/6) )/(4 (x)^(1/3)  − (x)^(1/(24)) )) ?
$$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{5}\:\sqrt[{\mathrm{3}}]{{x}}\:−\mathrm{3}\:\sqrt[{\mathrm{4}}]{{x}}\:−\sqrt[{\mathrm{6}}]{{x}}\:}{\mathrm{4}\:\sqrt[{\mathrm{3}}]{{x}}\:−\sqrt[{\mathrm{24}}]{{x}}}\:? \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{5}\:\sqrt[{\mathrm{3}}]{{x}}\:−\mathrm{3}\:\sqrt[{\mathrm{4}}]{{x}}\:−\sqrt[{\mathrm{6}}]{{x}}}{\mathrm{4}\:\sqrt[{\mathrm{3}}]{{x}}\:−\:\sqrt[{\mathrm{24}}]{{x}}}\:?\: \\ $$
Answered by bramlexs22 last updated on 08/Dec/20
(1) let x = t^(24)    lim_(t→0)  ((5t^8 −3t^6 −t^4 )/(4t^8 −t)) = lim_(t→0)  ((5t^7 −3t^5 −t^3 )/(4t^7 −1)) = 0
$$\left(\mathrm{1}\right)\:{let}\:{x}\:=\:{t}^{\mathrm{24}} \\ $$$$\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{5}{t}^{\mathrm{8}} −\mathrm{3}{t}^{\mathrm{6}} −{t}^{\mathrm{4}} }{\mathrm{4}{t}^{\mathrm{8}} −{t}}\:=\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{5}{t}^{\mathrm{7}} −\mathrm{3}{t}^{\mathrm{5}} −{t}^{\mathrm{3}} }{\mathrm{4}{t}^{\mathrm{7}} −\mathrm{1}}\:=\:\mathrm{0} \\ $$
Commented by bramlexs22 last updated on 08/Dec/20
(2)lim_(t→∞)  ((t^8 (5−(3/t^2 )−(1/t^4 )))/(t^8 (4−(1/t^7 )))) = (5/4)
$$\left(\mathrm{2}\right)\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\frac{{t}^{\mathrm{8}} \left(\mathrm{5}−\frac{\mathrm{3}}{{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{{t}^{\mathrm{4}} }\right)}{{t}^{\mathrm{8}} \left(\mathrm{4}−\frac{\mathrm{1}}{{t}^{\mathrm{7}} }\right)}\:=\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$

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