Limits of a Polynomial/Rational Function

[dekoi]

I understand that the limit as x -> a for a polynomial function, f(x), is equal to f(a) because the function is always continuous.

However, how can I prove this?

I also have to prove this for a rational function, including the fact that the denominator cannot equal to 0.

Thank you.

 

[amcavoy]

Ummm, you could try using the definition of a limit maybe:

\left|f(x)-L\right|<\epsilon

0<\left|x-a\right|<\delta

 

[Muzza]

What lemmas do you have available? I.e., do you know how to prove that the sum/product/quotient of two continuous functions is continuous?

 

[HallsofIvy]

By the time you get to the definition of "continuous function", you should already know:

1) lim_{x->a} c= c where c is a constant

2) lim_{x->a} x= a

3) If lim_{x->a}f(x)= L and lim_{x->a}g(x)= M then
lim_{x->a}f(x)+ g(x)= L + M

4) If lim_{x->a}f(x)= L and lim_{x->a}g(x)= M then
lim_{x->a}f(x)g(x)= LM

Since all polynomials consist of sums of products of x with itself and constants, it should be easy to use those to prove that lim_{x->a}P(x)= P(a) for P(x) any polynomial.

 

[dekoi]

Can you at least give me the first step?

I asked my professor and he also said to use the limit laws, but how can I begin?

Once I know the beginning I think I can figure out the rest on my own.

Thank you.

 

[CRGreathouse]

Can you at least give me the first step?

Decompose the polynomial into a collection of operations known to preserve continuity.

 

[dekoi]

Thank you. That has helped.

 

[HallsofIvy]

To show that a rational function is continuous every where except where the denominator is 0, you will need one more limit property:

If lim_{x->a}f(x)= L and lim_{x->a}g(x)= M then lim_{x->a}\frac{f(x)}{g(x)}= \frac{L}{M} provided M is not 0.