Question

Using the definition of limit to show that
lim x approaches -3, where 1-5x is under a surd and is equaling to 4. please help

Answer 1

you mean the
"\[\lim_{x\to a}f(x)=L\] if given any \(\epsilon >0\) there exists a \(\delta\) such that if \(|x-a|<\delta\) then \(|f(x)-L|<\epsilon\) definition?

Answer 2

yes, what I need to to is how exactly do you solve a problem under a surd using the defination.

Answer 3

you do these by working backwards
that is, write down exactly what you want, and then find the \(\delta\) in terms of \(\epsilon\) that gives it

first step is really to go from the general to the specific

Answer 4

you want
\[|\sqrt{1-5x}-4|<\epsilon\]

Answer 5

yes

Answer 6

lets see, maybe if write down what this means exactly, we can figure out how to achieve it

Answer 7

I did solve but am not really sure about my approach you know....

Answer 8

\[-\epsilon<\sqrt{1-5x}-4<\epsilon\]
\[4-\epsilon<\sqrt{1-5x}<4+\epsilon\] maybe a good start, although there might be a snappier way
in any case now we can get rid of the radical to get to the \(x\)

Answer 9

looks good
square all and get
\[16-8\epsilon+\epsilon^2<1-5x<16+8\epsilon+\epsilon^2
\\15-8\epsilon<-5x<15+8\epsilon\\
-3+\frac{8}{3}\epsilon>x>3-\frac{8}{3}\epsilon\]

Answer 10

typo there in the last line

Answer 11

\[-3+\frac{8}{3}\epsilon>x>-3-\frac{8}{3}\epsilon\]

Answer 12

now add 3 and put back in absolute value form
\[-\frac{8}{3}\epsilon<x+3<\frac{8}{3}\epsilon\]
\[|x+3|<\frac{8}{3}\epsilon\]

Answer 13

which is exactly what you want, since now you can say "let \(\delta=\frac{8}{3}\epsilon\) " and work the algebra backwards to prove that if this is the case, then
\[\sqrt{1-5x}-4|<\epsilon\] although usually you get to be done at that step

Answer 14

thanks hey, got distracted.

Answer 15

yw