# Continuous function.

1. Apr 19, 2004

### 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?

2. Apr 19, 2004

### master_coda

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.

3. Apr 19, 2004

### 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.

4. Apr 19, 2004

### 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?

5. Apr 20, 2004

### 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.