Continuous function.

[gimpy]

Im having a little trouble with this question.

If f is continuous at c and f(c) < 5, prove that there exists a \delta > 0 such that f(x) < 7 for all x \in (c - \delta , c + \delta)

So we are given that f is continuous at c.
So \lim_{x \to c}f(x) = f(c) < 5
\forall \epsilon > 0 \exists \delta > 0 such that whenever |x - c| < \delta then |f(x) - f(c)| < \epsilon

|x - c| < \delta
-\delta < x - c < \delta
c - \delta < x < c + \delta

Ok now im getting lost..
I know i have to do something with |f(x) - f(c)| < \epsilon
maybe
|f(x) - 5| < \epsilon
-\epsilon < f(x) - 5 < \epsilon
5 - \epsilon < f(x) < 5 + \epsilon
we want f(x) < 7 so .... am i on the right track??? How can i find the \delta > 0 that satisfies this?

[master_coda]

5 - \epsilon < f(x) < 5 + \epsilon
we want f(x) < 7 so .... am i on the right track??? How can i find the \delta > 0 that satisfies this?

Yes, you're on the right track. Take \epsilon=2. From the limit you mentioned, we know that there exists a \delta>0 such that 3<f(x)<7 and so you are done. The question was asking you to show that a delta exists, and the limit demonstrates that it clearly does, so you are done.

[gimpy]

I was actually thinking about making \epsilon = 2 but thought it was to easy. Ok so thanks for explaining the question a bit more to me, i understand it now. :smile:

[jdavel]

gimpy,

Isn't what you're asked to prove equivalent to,

\lim_{x \to c}f(x) = f(c) < 7

which follows trivially from,

\lim_{x \to c}f(x) = f(c) < 5

which was what you were given?

[HallsofIvy]

No, that's not what was given. Given that the limit is less than 5 it is true, but requires proof, that, for x close to c, f(x)< 5. gimpy was asked to prove the slightly simpler case: that, for x close to c, f(x)< 7.