f(x)=(x+1)^1/x. Calculate lim f(x); x->infinity.

f(x)=(x+1)^1/x.

Calculate lim f(x); x->infinity.

`lim_(x->oo) (x+1)^(1/x)`

` `This is in the indeterminate form infinity raised to zero. To find the limit we need to rewrite it into a from so that we can use L'Hopital's rule. 

Lets say that this limit is equal to L. Take the natural log of both sides: 

ln(L) = (1/x) ln(x +1)

Focusing only on the right side we get: 

`lim_(x->oo) ln(x+1)/(x)`

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` `Now we're in an infinity / infinity form and can use L'Hopital's rule. 

Take the derivative of the numerator and denominator to get: 

`lim_(x->oo) ((1)/(x+1))/(1)`

This limit is equal to 0. 

But we're not done yet!

Keep in mind that the original limit we're solving for is L. This 0 is actually ln(L). To solve for L, we need to undo the logarithm by using e. 

So L = e^0 = 1.