Find Delta...2

[nycmathguy]

Calculus
Section 2.4

This is not taught by Jenn in her videos. I need a detailed reply for 3 and 4. Your reply will be placed in my Calculus files to review when time allows. Thank you very much.

 

[MathLover1]

given graph of f(x)=sqrt(x) to find a number δ such that if |x-4|<δ then |sqrt(x)-2|<0.4

3.
|sqrt(x)−2|<0.4 is the same as −0.4<sqrt(x)−2<0.4

Since sqrt(x) is an increasing function, it is sufficient to look at the endpoints of the interval,
−0.4=sqrt(x)−2 and sqrt(x)−2=0.4.

For −0.4=sqrt(x)−2 we have immediately
sqrt(x) =−0.4+2
sqrt(x) =1.6
Squaring both sides, x=2.56.

For sqrt(x)−2=0.4
sqrt(x)=2+0.4
sqrt(x)=2.4 and
x=5.76

That is, | sqrt(x)−2|<0.4 is true as long as 2.56<x<5.76.

2.56=4−1.44
and
5.76=4+1.74

The smaller of those is 1.44 so |x−4|<1.44 will fit both.

4. do it same way

 

[nycmathguy]

given graph of f(x)=sqrt(x) to find a number δ such that if |x-4|<δ then |sqrt(x)-2|<0.4

3.
|sqrt(x)−2|<0.4 is the same as −0.4<sqrt(x)−2<0.4

Since sqrt(x) is an increasing function, it is sufficient to look at the endpoints of the interval,
−0.4=sqrt(x)−2 and sqrt(x)−2=0.4.

For −0.4=sqrt(x)−2 we have immediately
sqrt(x) =−0.4+2
sqrt(x) =1.6
Squaring both sides, x=2.56.

For sqrt(x)−2=0.4
sqrt(x)=2+0.4
sqrt(x)=2.4 and
x=5.76

That is, | sqrt(x)−2|<0.4 is true as long as 2.56<x<5.76.

2.56=4−1.44
and
5.76=4+1.74

The smaller of those is 1.44 so |x−4|<1.44 will fit both.

4. do it same way

I will do 4 tomorrow. Thank you.

See link below.

Do I rationalize the top or bottom as step 1?

Evaluate the Limit