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

Well, I know that $\lim_{x\rightarrow \infty}\left(1 + \dfrac{1}{x}\right)^x = e^1$

So I multipled the nominator and denominator of the exponent by $x$ so I can get closer to euler. but now I'm left to compute the limit of

${\dfrac{x^2y}{(x+y)x}} = {\dfrac{xy}{(x+y)}} = ?$

I tried polar, and I got $\dfrac{r\cos\phi \sin\phi}{\cos\phi + \sin\phi} = ?$

The question is, will the question mark be zero? because I know that the nominator is "zero". and I'm not sure about the denominator, I know that it is between $-2 \le x \le 2$. but if it is zero, I'm having a problem there. Maybe what I said is wrong and the denominator is just a number and makes altogether zero? $\frac{0}{k} = 0, k \in -2 \le x \le 2$ ?

  • 1
    $\begingroup$ Using polar coordinates here is a bad idea. $\endgroup$
    – Crostul
    Mar 10, 2016 at 13:30
  • $\begingroup$ Juste use the same method you use to prove that $\lim_{x\rightarrow \infty}\left(1 + \dfrac{1}{x}\right)^x = e^1$, that is, using $e$ and $\log$. $\endgroup$
    – Martigan
    Mar 10, 2016 at 13:31
  • $\begingroup$ @Crostul How would you approach this? $\endgroup$ Mar 10, 2016 at 13:32

2 Answers 2


It is much easier: $$\lim_{(x,y) \to (\infty , 1)} \frac{xy}{x+y} = \lim_{(x,y) \to (\infty , 1)} \frac{y}{1+y/x} = \frac{1}{1+0} = 1$$


If the limit exists then you can find its value by taking the limit along any path. In particular, the "limit as (x, y) goes to $(\infty, 1)$" can be taken by letting y go to infinity along the path y= 1. That is, take y= 1 first, so that the problem becomes $\lim_{x\to \infty} 2^{\frac{x^2}{x+ 1}}$.

That requires, as I said, knowing that the limit exists. The difficulty with using polar coordinates is that the line y= 1 does not go through the origin so is not easy to represent in polar coordinates. It might be simpler to change your "coordinate system", using the new variable, v= y- 1, so that the function becomes $\lim_{(x,v)\to (\infty, 0)} 2^{\frac{x(v- 1)}{xv- x}}$


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