Find: `lim_(x->oo)(1/(2 + 1/x))` and `lim_(x->oo-)(1/(2+1/x))`

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If you divide 1 by a number x, you will notice that as x takes on a larger value 1/x takes on a smaller value.

This continues for all values of x; if x1 > x2, (1/x1) < (1/x2). As x tends to infinity, 1/x tends to 0.

We have to find the value of `lim_(x->oo-)(1/(2 + 1/x))`. As x approaches infinity 1/x is equal to 0, this gives the value of the limit as 1/(2 + 0) = 1/2.

The value of `lim_(x->oo)(1/(2 +1/x))` is also equal to 1/2.

The required value of `lim_(x->oo-)(1/(2 +1/x))` = `lim_(x->oo)(1/(2+1/x))` = 1/2

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