# Limits of a Polynomial/Rational Function

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?

Homework Helper
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.

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.

Homework Helper
dekoi said:
Can you at least give me the first step?

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

Thank you. That has helped.

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.