- #1

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.

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- Thread starter dekoi
- Start date

- #1

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.

- #2

amcavoy

- 665

- 0

[tex]\left|f(x)-L\right|<\epsilon[/tex]

[tex]0<\left|x-a\right|<\delta[/tex]

- #3

Muzza

- 695

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- #4

HallsofIvy

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1) [tex]lim_{x->a} c= c[/tex] where c is a constant

2) [tex]lim_{x->a} x= a[/tex]

3) If [tex]lim_{x->a}f(x)= L[/tex] and [tex]lim_{x->a}g(x)= M[/tex] then

[tex]lim_{x->a}f(x)+ g(x)= L + M[/tex]

4) If [tex]lim_{x->a}f(x)= L[/tex] and [tex]lim_{x->a}g(x)= M[/tex] then

[tex]lim_{x->a}f(x)g(x)= LM[/tex]

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

- #5

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.

- #6

CRGreathouse

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

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

- #7

Thank you. That has helped.

- #8

HallsofIvy

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If [tex]lim_{x->a}f(x)= L[/tex] and [tex]lim_{x->a}g(x)= M[/tex] then [tex]lim_{x->a}\frac{f(x)}{g(x)}= \frac{L}{M}[/tex]

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MHB
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