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Question Number 77442 by aliesam last updated on 06/Jan/20
prove that lim_(x→∞) (∣((x^x^2 (x+2)^((x+1)^2 ) )/((x+1)^(2x^2 +2x+1) ))∣)=e
$${prove}\:{that} \\ $$$$\underset{{x}\rightarrow\infty} {{lim}}\:\left(\mid\frac{{x}^{{x}^{\mathrm{2}} } \left({x}+\mathrm{2}\right)^{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } }{\left({x}+\mathrm{1}\right)^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}} }\mid\right)={e} \\ $$
Answered by aliesam last updated on 06/Jan/20
L=lim_(x→∞) ∣((x^x^2 /((x+1)^x^2 ))) ×((((x+2)^((x+1)^2 ) )/((x+1)^((x+1)^2 ) )))∣ L=lim_(x→∞) (∣ ((x/(x+1)))^x^2 × (((x+2)/(x+1)))^((x+1)^2 ) ∣ L=lim_(x→∞) ∣( 1−(1/(x+1)))^x^2 × (1+(1/(x+1)))^((x+1)^2 ) ∣ L=lim_(x→∞) e^(∣x^2 ln(1−(1/(x+1)))+(x+1)^2 ln(1+(1/(x+1)))∣) lim_(z→0) L=lim_(x→∞) L when z=(1/(x+1)) and x^2 =(1/z^2 ) − (2/z) +1 (x+1)^2 =(1/z^2 ) L=lim_(z→0) e^(∣((ln(1−z))/z^2 ) − ((2ln(1−z))/z) + ln(1−z) +((ln(1+z))/z^2 )∣) L=e^(−1+2) =e
$${L}=\underset{{x}\rightarrow\infty} {{lim}}\:\:\mid\left(\frac{{x}^{{x}^{\mathrm{2}} } }{\left({x}+\mathrm{1}\right)^{{x}^{\mathrm{2}} } }\right)\:×\left(\frac{\left({x}+\mathrm{2}\right)^{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } }{\left({x}+\mathrm{1}\right)^{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } }\right)\mid\: \\ $$$${L}=\underset{{x}\rightarrow\infty} {{lim}}\left(\mid\:\left(\frac{{x}}{{x}+\mathrm{1}}\right)^{{x}^{\mathrm{2}} } ×\:\left(\frac{{x}+\mathrm{2}}{{x}+\mathrm{1}}\right)^{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \mid\:\right. \\ $$$${L}=\underset{{x}\rightarrow\infty} {{lim}}\mid\left(\:\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{{x}^{\mathrm{2}} } \:×\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \mid \\ $$$${L}=\underset{{x}\rightarrow\infty} {{lim}}\:{e}^{\mid{x}^{\mathrm{2}} {ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)+\left({x}+\mathrm{1}\right)^{\mathrm{2}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)\mid} \\ $$$$\underset{{z}\rightarrow\mathrm{0}} {{lim}}\:{L}=\underset{{x}\rightarrow\infty} {{lim}L}\:\:\:{when}\:{z}=\frac{\mathrm{1}}{{x}+\mathrm{1}}\:{and}\:{x}^{\mathrm{2}} =\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\:−\:\frac{\mathrm{2}}{{z}}\:+\mathrm{1} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{{z}^{\mathrm{2}} } \\ $$$$ \\ $$$${L}=\underset{{z}\rightarrow\mathrm{0}} {{lim}}\:{e}^{\mid\frac{{ln}\left(\mathrm{1}−{z}\right)}{{z}^{\mathrm{2}} }\:−\:\frac{\mathrm{2}{ln}\left(\mathrm{1}−{z}\right)}{{z}}\:+\:{ln}\left(\mathrm{1}−{z}\right)\:+\frac{{ln}\left(\mathrm{1}+{z}\right)}{{z}^{\mathrm{2}} }\mid} \\ $$$${L}={e}^{−\mathrm{1}+\mathrm{2}} ={e} \\ $$

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