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...
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.