Limit of 4x + 1 as x approaches 0 Okay, so I was approached with this question in my math class and I can answer the first two parts correctly but the third is throwing me off...
${f(x) = 4x + 1}$ 
a) Table of Values 
${x = -0.0001 , -0.001 , -0.01 , -0.1 , 0.1 , 0.01 , 0.001 , 0.0001}$
${f(x) = 0.9997, 0.997, 0.97, 0.7, 1.3, 1.03, 1.003, 1.0003}$
b) Make a conjecture about the value of limf(x) as x approaches zero 
1
c)Find an interval for x near 0 such that the difference between your conjectured limit and the value of the function is less than 0.01. (In other words, find  window of height 0.02 such that the graph exits the sides of the window and not the top or bottom of the window.)
____ ${< x <}$ _____

So, at first I thought that maybe if I subtracted the 0.9997 numbers from each of the 1.03 numbers that eventually I would get a value of 0.02 or at least a number really close to that value. All the conclusions I was arriving to, however, were incorrect. They were either too large or were 0.06 or 0.03 which I think is incorrect. 
 A: I would suggest simplifying your information to just what you want. Your conjectured value is 1, and and the actual value will be $4x+1$, so the difference will be
$$
error(x) = (4x + 1) - 1 = 4x.
$$
Your x-interval is going to be of the form $-\delta < x < \delta$ where $\delta > 0$. The errors of the $x$'s in this interval will be in the error interval
$$
error(-\delta) < error(x) < error(\delta)
$$
So if your goal is to have $-0.01 < error(x) < 0.01$, you just need to find the right $\delta$.
A: We can employ a simple $\delta,\epsilon$ proof for $f(x)=3x+1$ as $x\to0$.
Firstly, we can prove the conjecture proposed in part $\mathbb{b)}$: $\lim_{x\to0}f(x)=1$
That is, we want to show that $\forall \epsilon>0,\exists \delta>0, \ s.t.$
$$|x-a|<\delta\implies|f(x)-L|<\epsilon$$
Suppose $\lim_{x\to0}f(x)=1$, then for $\delta<\frac{\epsilon}{3}$:
$$|f(x)-L|=|3x+1-1|=|3x|<3\delta<\epsilon$$
$\therefore \lim_{x\to0}f(x)=1$.
We can extend this further to answer part $\mathbb{c)}$
We want to find a particular $\delta$ such that $-\delta<x<\delta\implies-0.01<(3x+1)-1<0.01$.
Well, we already found a general expression for $\delta$, in terms of the error $\epsilon$.
That is, let $\epsilon=0.01$, then clearly the above is satisfied when:
$3\delta<0.01\implies \delta<0.00\bar3=\frac {1}{300}$
There you have it!
