The function has a limit as f(x) = (1/x) + 3 has a limit of L=3 as x approaches x. This means that if x is sufficiently large (that is if x > N for some number N), the values of f(x) are closer to L=3 than a number epsilon > 0.

a) Sketch the graph y=(1/x) +3 and a horizontal strip of points (x,y) such that (if you are using the grapher, note that it has the capability to display more than one graph at once. The functions to be plotted are separated by a semicolon; you can input 1/x+3; 2.8; 3.2 for this problem). Based on the graph decide which portion of the graph is contained in the strip. Using inequalities the question can be reformulated as follows: If we want to guarantee that the values of f(x) = (1/x) +3 be closer to L=3; than epsilon = 0.2, that if require that |f(x)-3|<0.2, how large should the values of x be? That is what is a number N such that |f(x)-3|<epsilon if x > N?

(b) For the following values of epsilon , find N such that the graph y=(1/x) + 3 restricted to x>N is contained in the horizontal strip 3-epsilon<y<3+epsilon?

table to complete: |epsolon|0.2|0.@|0.05|0.001|
|N | | | | |

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a) Sketch the graph y=(1/x) +3 and a horizontal strip of points (x,y) such that (if you are using the grapher, note that it has the capability to display more than one graph at once. ...

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