Thread: epsilon delta question.

I read the great sticky on proving limits with epsilon and delta, but have a question on how to solve the problem below. I learned in class how to solve for epsilon with a function given, not just the graph :/. When i do have the function i understand i algebriaclly manipulate the equation (ie |(3x(x+1))/(x+1)| < epsilon to fit the equation of delta, ie |x-1| < delta.
After factoring and such i would conclude delta is |x-1|<epsilon/3, then i can just plug in any epsilon to get my delta. Heres the problem:

Use the given graph of f to find a number δ that fulfills the following condition.




im new to this, any help would be greatly appreciated, thank you.

But do you understand why that is the correct answer?

If you were given a slightly different graph, with different numbers, on a test, could you do the problem?

Notice that |f(x)- 3|< 0.6 is the same as -0.6< f(x)- 3< 0.6 and, adding 3 to each part, 3- 0.6= 2.4< f(x)< 3.6.

That's why the y axis has "2.4" and "3.6" marked as well as "3". Imagine drawing a horizontal line from a point on the y axis beween those two lines (representing some value of f(x) between 2.4 and 3.6) to the graph. Where you hit the graph, turn and draw a vertical line down, representing the x value that gives that f(x). That hits the x-axis between 4 and 5.7 and tells you that f(x) will be between 2.4 and 3.6 (so |f(x)- 3|< 0.6) as long as 4< x< 5.7. But you want so that which is the same as or so you want the same number on both sides. 4= 5-1 is 1 away from 5. 5.7= 5+ .7 is only .7 away. If you were to draw a vertical line 1 away from 5 but above it, at 6, that would be outside the two vertical lines. If you were to draw a vertical line .7 away from 5 but below it, at 5.3, that would be inside the two vertical lines. In order to stay "inside the lines" you need to use the smaller of the two differences which, here, is 0.7.