Problem: Find $\delta$ such that $x\to a$ with $0 < |x − a| < \delta$ implies that $|f (x) − L| < \epsilon$ where $\epsilon = 0.1$. (Use a graphing device)
$\lim\limits_{x\to -2} \sqrt{1-4x}=3$
Here are my steps:
$\lim\limits_{x\to -2} \sqrt{1-4x}=3$
$|\sqrt{1-4x}-3|<\frac{1}{10} $ ($|f(x)-L|<\varepsilon$)
$|\sqrt{1-4x}| < \frac{31}{10}$ (discard -3)
$|1-4x|<\frac{961}{100}$ (cancel the square root)
$|-4x|<\frac{861}{100}$
$|x| < \frac{861}{400}$
$|x-(-2)| < \frac{861}{400} -(-2)$ $\leftarrow|x-a|<\delta$
Take $\delta$ as $\frac{1661}{400}$(4.1525), while the standard answer is $\delta=\frac{59}{400}$.
Whether I did something wrong with the cancelling? Thanks in advance!
We want to find $\delta>0$ such that $|x+2|< \delta$ then $|\sqrt{1-4x}-3|<\frac1{10}$.
That is we want to ensure that
$$-\frac1{10}<\sqrt{1-4x}-3<\frac{1}{10}$$
$$3-\frac1{10}<\sqrt{1-4x}<3+\frac{1}{10}$$
$$(3-\frac1{10})^2<1-4x<(3+\frac{1}{10})^2$$
$$(3-\frac1{10})^2-1<-4x<(3+\frac{1}{10})^2-1$$
$$\frac{(3+\frac{1}{10})^2-1}{-4}<x<\frac{(3-\frac1{10})^2-1}{-4}$$
$$\frac{(3+\frac{1}{10})^2-1}{-4}+2<x+2<\frac{(3-\frac1{10})^2-1}{-4}+2$$
$$-0.1525<x+2<0.1475$$
Hence to make sure that the condition above happens,
we can choose $|x+2|<0.1475$.