# Thread: Limits in terms of epsilon and delta

1. ## Limits in terms of epsilon and delta

I hope this is clear. I am to find the value of a delta value > 0 such that |f(x) - L |< e when 0 < |x-a| < delta. I understand and can find the value of delta for limits such as lim -> -1 (2x-7) = -9, e = 0.1, delta = e/2 + .02/2 = 0.05. My problem is that I do not understand how to find delta > 0 for a limit x->2 x^3 = 8, e = .01. Can someone please show me how to write |f(x) - L|< e, -> |x^3-8|< .01. I tried using the difference of cubes |(x-2)(x^2 + 2x +4)|< .01. I do not know what to do from here, nor do I know if I set the absolute value less than .01 correctly. Can anybody help? The cube root and the square root functions are confusing me. Sorry, I couldn't insert the Greek letters in for epsilon nor delta.

2. Originally Posted by bosmith
I hope this is clear. I am to find the value of a delta value > 0 such that |f(x) - L |< e when 0 < |x-a| < delta. I understand and can find the value of delta for limits such as lim -> -1 (2x-7) = -9, e = 0.1, delta = e/2 + .02/2 = 0.05. My problem is that I do not understand how to find delta > 0 for a limit x->2 x^3 = 8, e = .01. Can someone please show me how to write |f(x) - L|< e, -> |x^3-8|< .01. I tried using the difference of cubes |(x-2)(x^2 + 2x +4)|< .01. I do not know what to do from here, nor do I know if I set the absolute value less than .01 correctly. Can anybody help? The cube root and the square root functions are confusing me. Sorry, I couldn't insert the Greek letters in for epsilon nor delta.
We should make use of the triangle inequality:

$\left|a+b\right|\leq \left|a\right|+\left|b\right|$.

By the definition of the limit,

$forall\,\varepsilon>0,\,\exists\,\delta>0:0<\left| x-2\right|<\delta$ whenever $\left|x^3-8\right|<\varepsilon$. Thus, if we find delta, then we have shown $\lim_{x\to 2}x^3=8$ to be true.

For us to continue, let us bound $\delta$, say $\delta=1$.

So,

$\left|x^3-8\right|=\left|(x-2)(x^2+2x+4)\right|=\left|x-2\right|\left|x^2+2x+4\right|$.

Let us find an upper bound on $\left|x^2+2x+4\right|$.

Applying the triangle inequality, we have

$\left|x^2+2x+4\right|=\left|x^2-4x+4+6x\right|\leq\left|(x-2)^2\right|+\left|6x-12+12\right|$ $=\left|x-2\right|^2+\left|6(x-2)+12\right|$ $\leq\left|x-2\right|^2+6\left|x-2\right|+\left|12\right|\leq 1^2+6(1)+12=19$

Thus, $\left|x-2\right|\left|x^2+2x+4\right|\leq 19\left|x-2\right|<\varepsilon\implies \left|x-2\right|<\frac{\varepsilon}{19}=\delta$.

So take $\delta=\frac{\varepsilon}{19}=\frac{.01}{19}=\frac {1}{1900}$

Does this make sense?