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Question Number 120515 by john santu last updated on 01/Nov/20

lim_(x→∞) (3 (2)^(1/(x )) −2.(3)^(1/(x )) )^x ?

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{3}\:\sqrt[{{x}\:}]{\mathrm{2}}\:−\mathrm{2}.\sqrt[{{x}\:}]{\mathrm{3}}\:\right)^{{x}} \:? \\ $$

Answered by bramlexs22 last updated on 01/Nov/20

ln L = lim_(x→∞) x ln (3.2^(1/x) −2.3^(1/x) ) let (1/x) = t ∧ t→0^+ ln L = lim_(t→0^+ ) ((ln (3.2^t −2.3^t ))/t) ln L = lim_(t→0^+ ) ((3.ln (2).2^t −2.ln (3).3^t )/(3.2^t −2.3^t )) ln L = ln ((8/9)). thus the limit is L=e^(ln ((8/9))) = (8/9)

$$\mathrm{ln}\:\mathrm{L}\:=\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}\:\mathrm{ln}\:\left(\mathrm{3}.\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{x}}} \:−\mathrm{2}.\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{x}}} \:\right) \\ $$$$\mathrm{let}\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{t}\:\wedge\:\mathrm{t}\rightarrow\mathrm{0}^{+} \\ $$$$\mathrm{ln}\:\mathrm{L}\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{3}.\mathrm{2}^{\mathrm{t}} −\mathrm{2}.\mathrm{3}^{\mathrm{t}} \right)}{\mathrm{t}} \\ $$$$\mathrm{ln}\:\mathrm{L}\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{3}.\mathrm{ln}\:\left(\mathrm{2}\right).\mathrm{2}^{\mathrm{t}} −\mathrm{2}.\mathrm{ln}\:\left(\mathrm{3}\right).\mathrm{3}^{\mathrm{t}} }{\mathrm{3}.\mathrm{2}^{\mathrm{t}} −\mathrm{2}.\mathrm{3}^{\mathrm{t}} } \\ $$$$\mathrm{ln}\:\mathrm{L}\:=\:\mathrm{ln}\:\left(\frac{\mathrm{8}}{\mathrm{9}}\right).\:\mathrm{thus}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{L}=\mathrm{e}^{\mathrm{ln}\:\left(\frac{\mathrm{8}}{\mathrm{9}}\right)} =\:\frac{\mathrm{8}}{\mathrm{9}} \\ $$

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