Dayaan M. answered • 9d
Earned A’s in Calc 1/AB & Calc 2/BC | 5 Years of Tutoring Experience
The short answer is no, you cannot guarantee it, and the reason is worth sitting with because it is the heart of how epsilon and delta actually work.
What you were actually given. The hypothesis says that for every x within 0.2 of 1, the value h(x) lands within 1 of 4. That is a single statement about a single fixed window. All it tells you is that h is trapped somewhere between 3 and 5 on the interval (0.8, 1.2). It does not say where in that band h sits, and it does not say h gets closer to 4 as x gets closer to 1.
What you are being asked for. You want a δ small enough that every x within δ of 1 forces h(x) within 0.5 of 4, meaning h trapped between 3.5 and 4.5. Shrinking δ shrinks the set of x values you are looking at, but nothing in the hypothesis makes the h values shrink toward 4 while you do it. That gap is the whole question.
The counterexample. Let h be the constant function h(x) = 4.9.
Check the hypothesis: |h(x) − 4| = 0.9 for every x, and 0.9 < 1, so whenever |x − 1| < 0.2 we do have |h(x) − 4| < 1. The given condition is satisfied.
Now check the conclusion: |h(x) − 4| = 0.9, and 0.9 is not less than 0.5, at any x whatsoever. So no matter how tiny you make δ, the conclusion fails for every single x in that window. No such δ exists, which is exactly what it means to say you cannot guarantee it.
Why this is the point of the exercise. A bound like "within 1" is a ceiling, not a promise that the values are near 4. The function is perfectly entitled to park itself at 4.9 and stay there. Choosing δ gives you control over which x values you inspect, and no control at all over what h does with them.
What would change the answer. If you were also told that h is continuous at x = 1 with h(1) = 4, or more directly that the limit of h(x) as x approaches 1 equals 4, then yes, such a δ would exist. That is precisely what the limit definition hands you: for every ε > 0, including ε = 0.5, there is some δ > 0 that works. The distance between that and what you were given is the distance between "h stays inside a band near 4" and "h actually approaches 4," and keeping those two ideas apart is the entire reason the epsilon delta definition is written the way it is.
If a picture helps more than symbols, draw the horizontal line y = 4.9 across the interval from 0.8 to 1.2, then shade the two bands, 3 to 5 and 3.5 to 4.5. The line lies inside the wide band and outside the narrow band everywhere along its length, and squeezing the x window inward from both sides does not move that line one bit.
