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**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now