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Possible Duplicate:
How to prove $(1+1/x)^x$ is increasing when $x>0$?

$f(x)=(1+1/x)^x$ Where $x>0$

I am in search to find a proof that the function $f(x)$ is always increasing in its any real number domain. As the above function always increasing a slight variation in the form of function will change the outcome in opposite way.That is when we change the exponent $x$ by ($1+x$) of the above function and letting all the expression on the right hand side intact, this new function will always be decreasing for real domain $x>0$.

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    Did you check the answers in [the link JavaMan posted](http://math.stackexchange.com/q/83035/856)?2012-11-25

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Here is an idea, it is based on the fact that the composition of two increasing functions is an increasing function. Let

$ f(x) = \left(1+\frac{1}{x}\right)^x \,. $

Now, consider the function

$ g(x) = x\ln\left(1+\frac{1}{x}\right) $

and prove that it is an increasing function. Then note that, $ f(x) = e^{g(x)} $ is a composition of two increasing functions ( since $e^x$ is an increasing function ).

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    @AbhinavAnand: In this case $g(x)=(1+x)\ln(1+1/x)$ which is a decreasing function.2012-11-25
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Compute the derivative. Are you familiar with first derivative test? The derivative will tell you the slope of the function at any given point, you can use this information to tell you about how the function behaves.

For example, positive slope means an increasing function.

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    I am extremely sorry that I overlooked your modest suggestion. The equation you mentioned and i arrived at after is a transcendental equation but still I reached at the proof through this.Don`t you have an email id.I have a wish to discuss this.2012-11-26