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So I want to solve that limit (why?) so here is my work so far: but professor says it is a false proof... So?

$$\lim_{x\to1}(\frac{4x^{101}}{x-1}+\frac{4}{1-x})=4/0+4/0=\infty$$

i also tried this: http://www.dummies.com/how-to/content/how-to-solve-limits-by-conjugate-multiplication.html

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    $\begingroup$ Note that the denominators differ by sign. It's $$\lim_{x\to 1} \frac{4(x^{101}-1)}{x-1}.$$ $\endgroup$ Jan 11, 2014 at 13:27
  • $\begingroup$ To answer the parentheses, well, there may be no specific reason for solving that limit other than your teacher wants you to, but limits are used for defining a lot of crazy things! $\endgroup$
    – GPerez
    Jan 11, 2014 at 13:32
  • $\begingroup$ Also to answer why it's a false proof, what your saying is that the limit of the sum is the sum of limits. This is ok... in some cases. Do you know which? $\endgroup$
    – GPerez
    Jan 11, 2014 at 13:35

2 Answers 2

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Using l'Hospital you get $$\lim_{x\to1} \frac{4x^{101} - 4}{x-1} = 4 \lim_{x\to1} \frac{x^{101}-1}{x-1} \stackrel{\text{l'H}}= 4\lim_{x\to1} \frac{101 x^{100}}1 = 404$$

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    $\begingroup$ I wouldn't bother with L'Hopital - this is the definition of the derivative of $x^{101}$ at $1$. $\endgroup$ Jan 11, 2014 at 14:14
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This can be done without l'Hopital's rule (although it is not computationally easier, it may be conceptually easier).

Note that $(x-1)$ is a factor of $(x^n-1)$ for integers $n\ge1$. For example:

$$(x-1)(x^4+x^3+x^2+x+1)=(x^5-1)$$

$$\therefore 4\lim_{x\rightarrow1}\dfrac{x^{101}-1}{x-1}=4\lim_{x\rightarrow1}\dfrac{x^{100}+x^{99}+x^{98}+\cdots+x^2+x+1}{1}$$

There are 101 terms in that second polynomial.

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