# Finding the target $\lim_{x\to1}(\frac{4x^{101}}{x-1}+\frac{4}{1-x})$

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

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

• Note that the denominators differ by sign. It's $$\lim_{x\to 1} \frac{4(x^{101}-1)}{x-1}.$$ Jan 11, 2014 at 13:27
• 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! Jan 11, 2014 at 13:32
• 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? Jan 11, 2014 at 13:35

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

• I wouldn't bother with L'Hopital - this is the definition of the derivative of $x^{101}$ at $1$. Jan 11, 2014 at 14:14

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.