Proving Continuity of f(x)/g(x) at c with Given Conditions

[kingstrick]

Want to show that f(x)/g(x) is continuous as x goes to c given that g(c) is not 0 and f(c) exists.

|f(x)/g(x) - f(c)/g(c)| = |1/(g(x)g(c))||f(x)g(c)-f(c)g(x)| = |1/(g(x)g(c))||f(x)g(c)-f(c)g(x)-f(x)g(x) + f(x)g(x)| <= |1/(g(x)g(c))|||f(x)||g(x)-g(c)| + |g(x)||f(x)- f(c)||

Now I am stuck

 

[Mark44]

Want to show that f(x)/g(x) is continuous as x goes to c given that g(c) is not 0 and f(c) exists.

|f(x)/g(x) - f(c)/g(c)| = |1/(g(x)g(c))||f(x)g(c)-f(c)g(x)| = |1/(g(x)g(c))||f(x)g(c)-f(c)g(x)-f(x)g(x) + f(x)g(x)| <= |1/(g(x)g(c))|||f(x)||g(x)-g(c)| + |g(x)||f(x)- f(c)||

Now I am stuck

You are missing some necessary conditions on f and g; namely, that they be continuous at c. You don't have any hope of proving that f/g is continuous if you don't have both f and g being continuous. It's not enough that g(c) is not 0 and f(c) exists.

 

[kingstrick]

I left that part out. Sorry I was trying to show where I was stuck. By staying f(x) is continuous at c, then can I state that f(x) -f(c) is zero by the definition of continuous?

 

[pasmith]

In the numerator, add and subtract $f(c)g(c)$ instead of $f(x)g(x)$.

 

[Mark44]

I left that part out. Sorry I was trying to show where I was stuck. By staying f(x) is continuous at c, then can I state that f(x) -f(c) is zero by the definition of continuous?

Depends on how rigorous you need to be. If you're not doing a δ-$\epsilon$ proof, then $lim_{x \to c} f(x) - f(c) = 0$, and I emphasize that this is a limit.

If you are doing a δ-$\epsilon$ proof, then |f(x) - f(c)| < $\epsilon_1$ and similar for |g(x) - g(c)|.

 

[kingstrick]

In the numerator, add and subtract $f(c)g(c)$ instead of $f(x)g(x)$.

That gives me |f(c)| |g(x) - g(c)| + |g(c)||f(x)-f(c)| in the numerator. Can I still claim that f(x) = f(c) ?

 

[HallsofIvy]

I left that part out. Sorry I was trying to show where I was stuck. By staying f(x) is continuous at c, then can I state that f(x) -f(c) is zero by the definition of continuous?

I don't know what you mean by that. Of course, f(x)- f(c) "is zero" at x= c. That is true because f(c)- f(c)= 0 whether f is continuous or not. Even if f is continuous, f(x)- f(c) is not necessarily equal to 0 for any other x. Perhaps you mean, not that f(x)- f(c) is 0, but that, for any $\epsilon> 0$, $|f(x)- f(x)|< \epsilon$ for x sufficiently close to c. That follows directly from the definition of "continuous", that $\lim_{x\to c} f(x)= f(c)$.

 

[kingstrick]

Depends on how rigorous you need to be. If you're not doing a δ-$\epsilon$ proof, then $lim_{x \to c} f(x) - f(c) = 0$, and I emphasize that this is a limit.

If you are doing a δ-$\epsilon$ proof, then |f(x) - f(c)| < $\epsilon_1$ and similar for |g(x) - g(c)|.

Thank you. We just learned about delta. I will try it that way.

 

[kingstrick]

Can I say

|f(x) - f(c)| <ϵepsilon/g(c)

since g(c) is a value and not equal to zero?

 

[kingstrick]

I am not sure how to handle the fraction

So now I have:

($\frac{1}{g(x)g(c)}$)|g(c)||f(x)-f(c)|+|f(c)||g(x)-g(c)|

given that |x-c|<$\delta$ then

($\frac{1}{g(x)g(c)}$)|g(c)||f(x)-f(c)|+|f(c)||g(x)-g(c)|=($\frac{1}{g(x)g(c)}$)|g(c)||x-c|+|f(c)||x-c|<$\epsilon$

so to find a good delta-epsilon i would need $\delta$ = $\epsilon$1 /|f(c)| + $\epsilon$2 /|g(c)| which i guess makes my $\epsilon$ = the summation of the two epsilons.

But this looks like i am going down the wrong rabbit hole and does nothing to show what i do with the fraction ($\frac{1}{g(x)g(c)}$)

 

[kingstrick]

**corrected

 

[PeroK]

You're perhaps missing a couple of tricks here:

First, there is no problem showing that f(x) gets close to f(c), so |f(x) - f(c)| gets small.

But, that's not what you want for g(x). Showing that g(x)-g(c) is small is not going to help. Instead, what you need is g(x) not to be too small. Because, if g(x) is too small, then f(x)/g(x) is going to get large and not close to f(c)/g(c).

So, the trick is show that |g(x)| is > |g(c)/2|, when x is close to c, which allows you to control the size of f(x)/g(x). To do this, you can chose e1 = |g(c)|/2 and show that as x gets close to c, |g(x) - g(c)| < |g(c)/2|, which means that |g(x)| > |g(c)|/2.

This is a common trick. Do you see the point?

 

[kingstrick]

If |g(x)| > |g(c)|/2 then 2/|g(c)| >1/|g(x)|

Allowing me to state 2/|g(c)g(c)| > 1/|g(c)g(x)|

 

[PeroK]

Yes.

Another idea is to prove that 1/g(x) is continuous. Then prove that the product of two functions is continuous. And, putting the two together, you have the result. This might split the difficulties up and make the thing more manageable. It is tricky!

 

[pasmith]

Yes.

Another idea is to prove that 1/g(x) is continuous.

Or just prove that 1/x is continuous where defined, and that a composition of continuous functions is continuous.

Then prove that the product of two continuous functions is continuous.

 

[kingstrick]

If it is given that g(c) exists and does not equal zero, Doesnt that mean 1/g(c) exists?

 

[kingstrick]

I guess I am asking if this is a iff statement. That if 1/g(c) exist implies that (1/g(x)) is continuous at c. Meaning that if g(x) is continuous thrn g(c) exists and given it is not 0 then 1/g(c) exists then 1/g(x) is continuous at c.

 

[PeroK]

You must assume:

f is continuous at c; g is continuous at c and g(c) not = 0.

Otherwise f/g may not be continuous at c.