I need help with this limit. I know the result is $e$ and I know I need to somehow modify the exponent into $x$ but am unsure as to how.

$\lim_{(x,y)\to(\infty, 1)}(1 + \frac{1}{x})^{\frac{x^2}{x+y}}$


  • $\begingroup$ as $x\to \infty$ we have $\frac{x^2}{x+1}\approx \frac{x^2}{x}=x$ it is $\lim x\to \infty \frac{\frac{x^2}{x}}{\frac{x}{x}+\frac{1}{x}}=x$ so we get $\lim_{x\to \infty} (1+\frac{1}{x})^x$ which is $e$ en.wikipedia.org/wiki/E_(mathematical_constant)#History $\endgroup$ – gbox Oct 17 '17 at 14:51
  • $\begingroup$ in second thought we may need to show both $\lim_{x\to \infty} \lim_{y\to 1}$ and $\lim_{y\to 1} \lim_{y\to \infty}$ the first one we showed the second one will be the same with $y$ as a constant so we get $\lim_{x\to \infty} \lim_{y\to 1}=\lim_{y\to 1} \lim_{y\to \infty}$ $\endgroup$ – gbox Oct 17 '17 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.