1. ## Limits: Find Delta

Hey, first post here and I just started this calculus I class.

Here's the question:

f(x) = 2 - (1/x)

Find Delta such that if 0 < |x - 1| < Delta then |f(x) - 1| < 0.1

Here's what I have so far:

|f(x) - 1|
= |2 - (1/x) - 1|
= |1 - (1/x)| (my book shows |f(x) - 1| = |(1/x) - 1|)

2. ## Re: Limits: Find Delta

Originally Posted by bearjew11
Here's the question:
f(x) = 2 - (1/x)
Find Delta such that if 0 < |x - 1| < Delta then |f(x) - 1| < 0.1
You must understand that $\displaystyle |a-b|=|b-a|$.
Then $\displaystyle \left| {\frac{1}{x} - 1} \right| = \frac{{\left| {x - 1} \right|}}{{\left| x \right|}}$

Now if $\displaystyle |x-1|<0.5$ then $\displaystyle \frac{1}{2}<x<\frac{3}{2}$ or $\displaystyle \frac{1}{{\left| x \right|}} < 2$

So pick $\displaystyle 0 < \delta < 0.05$.

3. ## Re: Limits: Find Delta

Originally Posted by bearjew11
Hey, first post here and I just started this calculus I class.

Here's the question:

f(x) = 2 - (1/x)

Find Delta such that if 0 < |x - 1| < Delta then |f(x) - 1| < 0.1

Here's what I have so far:

|f(x) - 1|
= |2 - (1/x) - 1|
= |1 - (1/x)| (my book shows |f(x) - 1| = |(1/x) - 1|)
Yes that's the same thing: |a- b|= |b- a|
Now you would want to write $\displaystyle \frac{1}{x}- 1= \frac{1}{x}- \frac{x}{x}= \frac{1- x}{x}$ and $\displaystyle \left|\frac{1- x}{x}\right|= \left|\frac{x- 1}{x}\right|$.

Now you can say "if $\displaystyle |f(x)- 1|< \epsilon$ then $\displaystyle \left|\frac{x- 1}{x}\right|< \epsilon$ and $\displaystyle |x- 1|< |x|\epsilon$ so you only need an upper bound on |x|. If x is "close to 1", say between 0 and 2, what is the largest |x| can be? What about $\displaystyle |x|\epsilon$?