Question

hi mathematicians,for those who are doing advanced calculus(limits).can u help me with this:let p(x) be a polynomial function and a an element of real numbers.PROVE THAT:limp(x) as x turns to a is equal to p(a).

Answer 1

ok for one thing a polynomial is continuous, but i guess that is what you are trying to prove using "limit laws"
you will need three of them

Answer 2

\[\lim_{x\to a}x^n=a^n\]
\[\lim_{x\to a} cf(x)=c\lim_{x\to a} f(x)\] that is, a constant does not effect the limit, and finally
\[\lim_{x\to a}(f(x)+g(x))=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)\] that is a limit is linear

Answer 3

so you can write a polynomial as
\[p(x)=c_nx^n+c_{n-1}x^{n-1}+ ... + c_1xc+c_0\]
and you want
\[\lim_{x\to a}p(x)=p(a)\] so write
\[\lim_{x\to a}p(x)=\lim_{x\to a}c_nx^n+c_{n-1}x^{n-1}+ ... + c_1xc+c_0\] then use the limit laws one at a time

Answer 4

for example, the last law allows you to write this as
\[\lim_{x\to a}c_nx^n+\lim_{x\to a}c_{n-1}x^{n-1}+ ... + \lim_{x\to a}c_1xc+\lim_{x\to a}c_0\]

Answer 5

ok,i understand you but i dont think we should use limit laws.how about using the definition of a limit.

Answer 6

is'nt that,p(a)=L and p(x)=f(x).so we have a related function

Answer 7

as satellite pointed out, you can expand the polynomial as such:
\[p(x)=c_nx^n+c_{n-1}x^{n-1}+...+c_2x^2+c_1x+c_0\]
as such,
\[p(a)=c_na^n+c_{n-1}x^{n-1}+...+c_2a^2+c_1a+c_0\]
\[\lim_{x\rightarrow a} p(x)=\lim_{x\rightarrow a}c_nx^n+\lim_{x\rightarrow a}c_{n-1}x^{n-1}+...+\lim_{x \rightarrow a}c_1x+\lim_{x\rightarrow a}c_0\]
etc. that should be a big hint to you :D