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What is the following limit? $$ \lim_{x \to \infty} {\displaystyle{\large\left(1 + x\right)\left(1 + 1/x\right)^{2x} - x\,{\rm e}^{2}} \over \displaystyle{\large{\rm e}^{2} - \left(1 + 1/x\right)^{2x}}} $$ Thanks.

Felix Marin
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user156001
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    I think you're supposed to show what you've done so far. – PA6OTA Jun 17 '14 at 22:58
  • Welcome !!!. You are a new user. M.SE people think you must show what did you try to get the answer and/or where you stop to see the road to the answer. In that way they can provide you with several alternatives. – Felix Marin Jun 17 '14 at 23:02
  • This is a nice hard limit to do; the limit exists ($\frac{1}{6}$ seems to be the answer) but none of the usual tricks nail it. – Mark Fischler Jun 18 '14 at 00:09

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Define the numerator and the denominator of the limit as:

$$a(x):={\large\left(1 + x\right)\left(1 + 1/x\right)^{2x} - x\,{\rm e}^{2}}$$ $$b(x):={\large{\rm e}^{2} - \left(1 + 1/x\right)^{2x}}$$

Then $$A(t)=a(1/t)={\large\left(1 + (1/t)\right)\left(1 + t\right)^{2/t} - (1/t)\,{\rm e}^{2}}$$

$$B(t)=b(1/t)={\large{\rm e}^{2} - \left(1 + t\right)^{2/t}}$$

We can now expand $A(t)$ and $B(t)$ at $t=0$ ($x=\infty$):

$$A(t)=\frac{{\rm e}^{2}}{6}t+O(t^2)$$

$$B(t)={\rm e}^{2}t+O(t^2)$$

Thus $$\lim_{x \to \infty}\frac{a(x)}{b(x)}=\lim_{t \to 0}\frac{A(t)}{B(t)} =\lim_{t \to 0}\frac{\frac{{\rm e}^{2}}{6}t}{{\rm e}^{2}t}=\frac{1}{6}$$

It is exactly as what Mark Fischler predicted.

mike
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    I really like that trick of converting the problem to limit at zero: "We can now expand A(t) and B(t) at t=0 (x=∞)" – DWin Jun 18 '14 at 00:52