Problem:

let f(x)=sqrt(19-x) and lim x-->10 f(x)=3 . Use the graph (or table) to find a value delta whenever epsilon=.5

Can someone please help me with this problem? Thanks

I don't even know where to begin...

Printable View

- Jun 21st 2009, 02:58 PMjohnny4lsuNeed help finding delta....
Problem:

let f(x)=sqrt(19-x) and lim x-->10 f(x)=3 . Use the graph (or table) to find a value delta whenever epsilon=.5

Can someone please help me with this problem? Thanks

I don't even know where to begin... - Jun 21st 2009, 03:33 PMBruno J.
You are looking for a value of $\displaystyle \delta$ such that whenever $\displaystyle |x-10|<\delta$ then $\displaystyle |f(x)-f(10)|<\epsilon$.

Where do you get stuck? - Jun 21st 2009, 03:38 PMVonNemo19

This is how I do this

But, your problem says to find the value by lookind at the graph, which is much easier. I'll talk about that in a minute.

$\displaystyle f(x)=\sqrt{19-x}$

$\displaystyle \lim_{x\to{10}}\sqrt{19-x}=3$

We wish to find an $\displaystyle \delta$ such that $\displaystyle |\sqrt{19-x}-3|<0.5$ whenever $\displaystyle |x-10|<\delta$

So here we just solve the double inequality

$\displaystyle -0.5<\sqrt{19-x}-3<0.5$

$\displaystyle 2.5<\sqrt{19-x}<3.5$

squaring (note that the direction of the signs is preserved)

$\displaystyle 6.25<19-x<12.25$

$\displaystyle -12.75<-x<-6.75$

$\displaystyle 12.75>x>6.75$

So now we need to look at

$\displaystyle |x-10|<\delta$

So we choose $\displaystyle \delta=2.75$

Analyling limits graphically, although imprecise, is easy.

1.Graph the function

2. graph the lines y=3, y=3.5, and y=2.5

3. where the lines y=3.5 and y=2.5 intersect the graph of the function, draw vertical lines.

x=10 will be in the interval where these lines cross. Approximate delta by choosing delta just a little bit smaller than the absolute value of the difference between 10 and the aproximate value of x where the vertical line closest to x=10 crosses the x-axis. this is your delta. - Jun 21st 2009, 04:00 PMjohnny4lsu
thanks a bunch. that way of doing it makes much more sense to me. Thanks again!