What is the limit of (x/x+1)^x as x tends to infinity?

[l33t_V]

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

 

[Mark44]

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).

 

[g_edgar]

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

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

 

[l33t_V]

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).

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

 

[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

 

[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}