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Question Number 211750 by Alijumaaxyz last updated on 19/Sep/24
lim_(x→∞) (((a^(1/x) +b^(1/x) )/2))^x ;(a,b)∈R_+ ^∗
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{a}^{\mathrm{1}/{x}} +{b}^{\mathrm{1}/{x}} }{\mathrm{2}}\right)^{{x}} ;\left({a},{b}\right)\in\mathbb{R}_{+} ^{\ast} \\ $$
Answered by mehdee7396 last updated on 19/Sep/24
lim_(x→∞) b(((((a/b))^(1/x) +1)/2))^x ★ lim_(x→∞) (((((a/b))^(1/x) +1)/2))^x =lim_(x→∞) (((((a/b))^(1/x) +1)/2)−1)x =lim_(x→∞) (((((a/b))^(1/x) −1)/2))x =lim_(x→∞) ((((a/b))^(1/x) −1)/(2/x)) =lim_(x→∞) ((−(1/x^2 )((a/b))^(1/x) ln((a/b)))/(−(2/x^2 )))=(1/2)ln((a/b)) ⇒lim_(x→∞) (((((a/b))^(1/x) +1)/2))^x =e^((1/2)ln((a/b))) =(√(a/b)) ⇒lim_(x→∞) b(((((a/b))^(1/x) +1)/2))^x = (√(ab)) ✓
$${li}\underset{{x}\rightarrow\infty} {{m}}\:{b}\left(\frac{\left(\frac{{a}}{{b}}\right)^{\frac{\mathrm{1}}{{x}}} +\mathrm{1}}{\mathrm{2}}\right)^{{x}} \\ $$$$\bigstar\:{li}\underset{{x}\rightarrow\infty} {{m}}\:\left(\frac{\left(\frac{{a}}{{b}}\right)^{\frac{\mathrm{1}}{{x}}} +\mathrm{1}}{\mathrm{2}}\right)^{{x}} \\ $$$$={li}\underset{{x}\rightarrow\infty} {{m}}\:\left(\frac{\left(\frac{{a}}{{b}}\right)^{\frac{\mathrm{1}}{{x}}} +\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right){x} \\ $$$$={li}\underset{{x}\rightarrow\infty} {{m}}\:\left(\frac{\left(\frac{{a}}{{b}}\right)^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}}{\mathrm{2}}\right){x} \\ $$$$={li}\underset{{x}\rightarrow\infty} {{m}}\:\frac{\left(\frac{{a}}{{b}}\right)^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}}{\frac{\mathrm{2}}{{x}}} \\ $$$$={li}\underset{{x}\rightarrow\infty} {{m}}\:\frac{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\frac{{a}}{{b}}\right)^{\frac{\mathrm{1}}{{x}}} {ln}\left(\frac{{a}}{{b}}\right)}{−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{a}}{{b}}\right) \\ $$$$\Rightarrow{li}\underset{{x}\rightarrow\infty} {{m}}\:\left(\frac{\left(\frac{{a}}{{b}}\right)^{\frac{\mathrm{1}}{{x}}} +\mathrm{1}}{\mathrm{2}}\right)^{{x}} ={e}^{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{a}}{{b}}\right)} =\sqrt{\frac{{a}}{{b}}} \\ $$$$\Rightarrow{li}\underset{{x}\rightarrow\infty} {{m}}\:{b}\left(\frac{\left(\frac{{a}}{{b}}\right)^{\frac{\mathrm{1}}{{x}}} +\mathrm{1}}{\mathrm{2}}\right)^{{x}} =\:\sqrt{{ab}}\:\:\checkmark \\ $$$$ \\ $$
Commented by Alijumaaxyz last updated on 19/Sep/24
✓ thank you sir
$$\checkmark\:{thank}\:{you}\:{sir} \\ $$

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