anonymous
  • anonymous
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).
Mathematics
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chestercat
  • chestercat
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anonymous
  • anonymous
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
anonymous
  • anonymous
\[\lim_{x\to a}x^n=a^n\] \[\lim_{x\to a} cf(x)=c\lim_{x\to a} f(x)\] 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
anonymous
  • anonymous
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

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anonymous
  • anonymous
for example, the last law allows you to write this as \[\lim_{x\to a}c_nx^n+\lim_{x\to a}c_{n-1}x^{n-1}+ ... + \lim_{x\to a}c_1xc+\lim_{x\to a}c_0\]
anonymous
  • anonymous
ok,i understand you but i dont think we should use limit laws.how about using the definition of a limit.
anonymous
  • anonymous
is'nt that,p(a)=L and p(x)=f(x).so we have a related function
agreene
  • agreene
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

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