KCC's quizzes AQQ255 about a limit at infinite

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

Fantastic Tim! You get it, congratulations!