# limit x to infinity (1+1/6x)^4x+1

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### 1 Answer

You need to substitute oo for x in equation under limit such that:

`lim_(x-gtoo) (1+1/6x)^(4x+1) = (1+1/oo)^(oo)`

`lim_(x-gtoo) (1+1/6x)^(4x+1) = (1+0)^(oo) = 1^oo`

Notice that the limit is indeterminate, hence you need to create the special limit `lim_(x-gtoo)(1 + 1/x)^x = e` such that:

`lim_(x-gtoo) ((1+1/6x)^(6x))^((4x+1)/(6x)) `

Notice that `lim_(x-gtoo) ((1+1/6x)^(6x)) = e` , hence, evaluating `lim_(x-gtoo) ((1+1/6x)^(6x))^((4x+1)/(6x))` yields `e^(lim_(x-gtoo) ((4x+1)/(6x)))` `lim_(x-gtoo) (4x+1)/(6x)= lim_(x-gtoo) (x(4+1/x))/(6x)`

Reducing by x yields:

`lim_(x-gtoo) (4+1/x)/6 =lim_(x-gtoo) 4/6 + lim_(x-gtoo) 1/(6x)`

`lim_(x-gtoo) (4+1/x)/6 = 2/3 + 0`

`lim_(x-gtoo) ((1+1/6x)^(6x))^((4x+1)/(6x)) = e^(2/3)`

`lim_(x-gtoo) ((1+1/6x)^(6x))^((4x+1)/(6x)) = root(3)(e^2)`

**Hence, evaluating the limit of the function yields `lim_(x-gtoo) ((1+1/6x)^(6x))^((4x+1)/(6x)) = root(3)(e^2).` **