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KCC's quizzes AQQ255 about a limit at infinite

And here is a new quiz coming!

1. Info: the frequent winners should have received some award and certificate. Please warn me if you think you have been forgotten!

2. Quote of the week: "The elevator to success is out of order. You'll have to use the stairs... one step at a time." - Joe Girard.

3. New quiz of the week: AQQ255 about a limit to infinite

Good luck!

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[edited by: emassa at 3:52 PM (GMT -4) on 21 Mar 2024]
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• 1)  It is the math constant 'e'

2)  y=\lim_{x\to\text{infinity}}(1+\frac{1}{x})^x

\ln(y)=\lim_{x\to\text{infinity}}x\ln(1+\frac{1}{x})

rearrange to:

\ln(y)=\lim_{x\to\text{infinity}}\ln\frac{(1+\frac{1}{x})}{\frac{1}{x}}

This limit leads to 0/0, so we apply l'Hopital's rule and take derivative

\frac{-\frac{1}{x^2}}{\frac{(1+\frac{1}{x})}{-\frac{1}{x^2}}}

Take the limit again, which gives:

\ln(y)=\lim_{x\to\imaginaryI nf}(1+\frac{1}{x})

\ln(y)=1

e^{\ln(y)}=e^1

y=e

• 1)  It is the math constant 'e'

2)  y=\lim_{x\to\text{infinity}}(1+\frac{1}{x})^x

\ln(y)=\lim_{x\to\text{infinity}}x\ln(1+\frac{1}{x})

rearrange to:

\ln(y)=\lim_{x\to\text{infinity}}\ln\frac{(1+\frac{1}{x})}{\frac{1}{x}}

This limit leads to 0/0, so we apply l'Hopital's rule and take derivative

\frac{-\frac{1}{x^2}}{\frac{(1+\frac{1}{x})}{-\frac{1}{x^2}}}

Take the limit again, which gives:

\ln(y)=\lim_{x\to\imaginaryI nf}(1+\frac{1}{x})

\ln(y)=1

e^{\ln(y)}=e^1

y=e

Children