Epsilon-Delta Limits: Finding the Optimal Delta for a Given Epsilon Value

[ƒ(x)]

Given: limit of (sin x)/x as x --> 0 and that ε = .01
Problem: Find the greatest c such that δ between zero and c is good. Give an approximation to three decimal places.

Equations:

0 < |x - a| < δ
0 < |f(x) - L| < εAttempt:

0 < |x - 0| < δ
0 < | sin(x)/x - 1| < ε

0 < | sin(x)/x - 1| < .01
0 < | sin(x)/x| < 1.01
0 < |sin(x)| < 1.01|x|
0 < |sin(x)| < 1.01δ
0 < |sin(x)|/1.01 < δ
Since sin(x) is going to range between -1 and 1, the greatest value for δ is 1/1.01. But, this answer isn't correct. The correct answer is .245, and I don't know how to get that.

 

[jbunniii]

Given: limit of (sin x)/x as x --> 0 and that ε = .01
Problem: Find the greatest c such that δ between zero and c is good. Give an approximation to three decimal places.

What do you mean by "δ between zero and c is good"?

My guess is that you mean:

"Find the greatest c such that

\left|\frac{\sin x}{x} - 1 \right| &lt; \epsilon

whenever

0 &lt; |x - 0| &lt; \delta \leq c."

Is that right?

[It seems clearer to dispense with c and simply ask "how big can \delta be?"]

Your problem is the following step:

Equations:

0 < | sin(x)/x - 1| < .01
0 < | sin(x)/x| < 1.01

While the upper bound in the second line is true, it doesn't buy you anything. In fact, it's true no matter what \epsilon is, because |\sin(x)/x| never exceeds 1.

Also, the lower bounds are wrong: both |\sin(x)/x - 1| and |\sin(x)/x| do equal 0 for some values of x. We don't want/need to disallow that possibility.

Try writing

\left|\frac{\sin x}{x} - 1\right| &lt; 0.01

in the following equivalent way

-0.01 &lt; \frac{\sin x}{x} - 1 &lt; 0.01

and thus

0.99 &lt; \frac{\sin x}{x} &lt; 1.01

The right-hand inequality is true for all x, so we only need to consider the left inequality:

0.99 &lt; \frac{\sin x}{x}

Now you need to find what range of x satisfies this inequality. I assume you are permitted to do so numerically (using a computer or calculator)?

 

[ƒ(x)]

Yes, that works. Thank you.

 

[jbunniii]

Oops, for the record, I just noticed that I lied when I said that |\sin(x)/x - 1| can equal 0. It can't. But my point that you don't need the 0 &lt; |\sin(x)/x - 1| part of the inequality is still correct.

 

[ƒ(x)]

Ok. :)