# Lim (x/x+1)^x

1. Oct 16, 2009

### l33t_V

Can someone help me with finding the limit of (x/x+1)^x as x tends to infinity

2. Oct 16, 2009

### Staff: Mentor

The limit is infinity, unless of course this is what you meant:
$$\lim_{x \rightarrow \infty} \left ( \frac{x}{x + 1}\right )^x$$

If you want to write the quotient of x and x + 1, put parentheses around what goes in the denominator, like so: x/(x + 1).

3. Oct 16, 2009

### g_edgar

The reciprocal is
$$\lim_{x \rightarrow \infty} \left ( \frac{x+1}{x}\right )^x$$
Can you do that limit?

4. Oct 16, 2009

### l33t_V

Yes, i meant (x/(x+1))^x

Last edited: Oct 16, 2009
5. Oct 16, 2009

### l33t_V

Never mind, I solved it.

(x/(x(1+1/x)))^x = (1/(1+1/x))^x = 1^x/(1+1/x)^x where 1^x = 1 and (1+(1/x))^x as x tends to inf = e^1

therefore as x tends to infinity the function tends to e^-1

6. Oct 18, 2009

### centry57

$$\lim_{x\rightarrow \infty} (x/x+1)^x= \lim _{x\rightarrow \infty} (1+\frac{-1}{x+1})^{-(x+1)\frac{x}{-(x+1)}})=\lim_{x\rightarrow \infty} e^\frac{x}{-(x+1)}}=e^{-1}$$