1. Mar 23, 2005

### castroltran

I stumble on this problem that I have no clue how to do. Can someone please help me. I was thinking that this problem might be somehow can be solve by taking the natural log and then e back again. But I am stuck somehow.

2. Mar 23, 2005

### dextercioby

Make the substitution

$$\frac{2}{x}=\frac{1}{u}$$

Daniel.

3. Mar 23, 2005

### DH

lim(1+2/x)^x when x -> infinity
ok,
At first: 1 + 2/x = (x + 2)/x
You set:
y = lim(1+2/x)^x when x -> infinity
-> lny = ln (lim(1+2/x)^x when x -> infinity)
Since ln(lim(1+2/x)^x when x -> infinity) is continuous function
-> lny = lim[ln(1+2/x)^x when x -> infinity]
-> lny = lim[xln(1+2/x) when x -> infinity]
-> lny = lim[ln(1+2/x)/(1/x)]
We can see that this limit has form: infinity/infinity
-> We use L'Hopital rule, take derivative:
derivative of ln(1+2/x) = (-2/x^2)/(1+2/x) = -2/[x(x+2)]
derivative of 1/x = -1/x^2
-> lny = lim [2x/(x+2)] when x -> infinity
One more time use L'Hopital rule
-> lny = lim (2/1) = 2
-> y = e^2
e^2 = lim(1+2/x)^x when x -> infinity

4. Mar 23, 2005

### castroltran

thank you so much, you have just saved my life. Thank you for the detail explaination =)

5. Mar 23, 2005

### dextercioby

$$x=2u$$

$$\lim_{x\rightarrow +\infty} \left(1+\frac{2}{x}\right)^{x}=\lim_{u\rightarrow +\infty} \left(1+\frac{1}{u}\right)^{2u}=e^{2}$$

Daniel.