I have to solve the following limit:

\begin{align*} \lim_{\substack{x\to\infty\\y\to0\,\,}}\left(1+\frac{1}{x}\right)^{\frac{x^2}{x+y}} &= \lim_{\substack{x\to\infty\\y\to0\,\,}}\left\{\left[\left(1+\frac{1}{x}\right)^x\;\right]^\frac{x^2}{x+y}\right\}^\frac{1}{x}\\ &= \lim_{\substack{x\to\infty\\y\to0\,\,}} e\cdot\frac{x}{x+y}\\ &=e \end{align*} Paint.

So, this is the solution I got. Is everything correct?

.....And something else. I really don't have a clue about this one:

$$ \lim_{\substack{x\to\infty\\y\to\infty}}\frac{\exp\left\{-\dfrac{1}{x^2+y^2}\right\}}{x^4+y^4}.$$ Paint2.

Can you guys help me find a solution?



closed as unclear what you're asking by John B, user228113, Claude Leibovici, choco_addicted, Watson Mar 17 '16 at 7:46

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ I appreciate the effort, but please, don't every use paint again. Review the edit. Formatting tips here. $\endgroup$ – Em. Mar 16 '16 at 19:05
  • $\begingroup$ Got it! Thanks! $\endgroup$ – snitchben Mar 16 '16 at 19:06
  • $\begingroup$ Also, I used $\exp$ instead of $e$ for legibility. As Jan writes $\exp (x) = e^x$. $\endgroup$ – Em. Mar 16 '16 at 19:07
  • $\begingroup$ Thanks probablyme! Do you have any idea on the second limit? $\endgroup$ – snitchben Mar 16 '16 at 19:09
  • $\begingroup$ It's ok haha. I wish I didn't have to do calculus too $\endgroup$ – snitchben Mar 16 '16 at 19:12

$$\lim_{x\to\infty}\left[\lim_{y\to0}\left[\left(1+\frac{1}{x}\right)^{\frac{x^2}{x+y}}\right]\right]=\lim_{x\to\infty}\left[\lim_{y\to0}\left[\exp\left[\ln\left(\left(1+\frac{1}{x}\right)^{\frac{x^2}{x+y}}\right)\right]\right]\right]=$$ $$\lim_{x\to\infty}\left[\lim_{y\to0}\left[\exp\left[\frac{x^2\ln\left(1+\frac{1}{x}\right)}{x+y}\right]\right]\right]=\lim_{x\to\infty}\left[\exp\left[\lim_{y\to0}\frac{x^2\ln\left(1+\frac{1}{x}\right)}{x+y}\right]\right]=$$ $$\lim_{x\to\infty}\exp\left[\frac{x^2\ln\left(1+\frac{1}{x}\right)}{x+0}\right]=\exp\left[\lim_{x\to\infty}\frac{x^2\ln\left(1+\frac{1}{x}\right)}{x}\right]=$$ $$\exp\left[\lim_{x\to\infty}\frac{x\ln\left(1+\frac{1}{x}\right)}{1}\right]=\exp\left[\lim_{x\to\infty}x\ln\left(1+\frac{1}{x}\right)\right]=$$

Applying l'Hôpital's rule, to get:


  • $\begingroup$ Thanks Jan, but I think I haven't learnt this method yet. I don't know what exp is, so I am not allowed to use it $\endgroup$ – snitchben Mar 16 '16 at 19:03
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    $\begingroup$ @snitchben $\exp(x)=e^x$ $\endgroup$ – Jan Mar 16 '16 at 19:04

Hint for the second one: The expression is $\le 1/(x^4+y^4).$

  • $\begingroup$ Thanks zhw., but I don't understand what you mean $\endgroup$ – snitchben Mar 16 '16 at 19:12
  • $\begingroup$ If $a\ge 0,$ then $e^{-a}\le 1.$ Also, please answer my question above. $\endgroup$ – zhw. Mar 16 '16 at 19:38

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