1
$\begingroup$

I know this is probably a really stupid question but I am stuck.

$$\lim_{x\to -2} \frac{x^{2n+1} + 2^{2n+1}}{x+2}: n \in \mathbb{J}^+$$.

$\mathbb{J}^+$ being the positive algebraic subset of irrational.

I know this is probably ridiculously bad formatting but I really don't know how to write it.

  • 0
    http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference will help. Also, J+? Do you mean $\Bbb N$ instead or $\Bbb Z^+$? I've never heard of a set of numbers being referred to as J before... – 2017-02-04
  • 0
    I edited the question to add proper formatting. If I have misinterpreted something, please let me know or edit it again. Regarding the actual mathematics, think about the derivative of the function $x\mapsto x^{2n+1}$, and what its value is at $x=-2$. What you wrote is an incremental quotient. – 2017-02-04

3 Answers 3

2

$$\frac{x^{2n+1}+2^{2n+1}}{x+2}=\frac{\require{cancel}\cancel{(x+2)}(x^{2n}-2x^{2n-1}+2^2x^{2n-2}-\ldots+2^{2n})}{\cancel{x+2}}\xrightarrow[x\to-2]{}\ldots$$

(Watch carefully the sign of each summand...)

Easier. With l'Hospital:

$$\lim_{x\to-2}\frac{x^{2n+1}+2^{2n+1}}{x+2}\stackrel{l'H}=\lim_{x\to-2}(2n+1)x^{2n}=(2n+1)2^{2n}$$

  • 0
    I am so sorry, could you expand on the steps during l'Hospital I am still lost between there. – 2017-02-04
  • 0
    @TEKNO Just differentiate numerator and denominator...with respect to $\;x\;$ , of course. – 2017-02-04
  • 0
    really sorry to bother you again but am I doing something wrong here? http://imgur.com/a/E1vJ0 – 2017-02-04
  • 0
    @TEKNO Yes, you are: you are differentiating $\;2^{2n+1}\;$ and getting $\;(2n+1)2^n\;$ ...but this is a constant number! Its derivative is zero...The only variable here is $\;x\;$ . – 2017-02-04
1

It's not difficult to prove that, for any $a,b\in\mathbb{R}$ and $n\in\mathbb{N}$ :

$$a^{2n+1}+b^{2n+1}=(a+b)\sum_{k=0}^{2n}(-1)^{k}a^kb^{2n-k}$$

Hence, if $x\neq 2$ :

$$\frac{x^{2n+1}+2^{2n+1}}{x+2}=\sum_{k=0}^{2n}(-1)^k2^kx^{2n-k}$$and so :

$$\lim_{x\to-2}\frac{x^{2n+1}+2^{2n+1}}{x+2}=\sum_{k=0}^{2n}(-1)^k2^k(-2)^{2n-k}=(2n+1)2^{2n}$$

0

$$\frac{x^{2n+1} + 2^{2n+1}}{x+2}$$

Factor out $2^{2n+1}$ on top and $2$ on bottom,

$$2^{2n}\frac{(x/2)^{2n+1}+1}{(x/2)+1}$$

$$=2^{2n}\frac{1-(-x/2)^{2n+1}}{1-(-x/2)}$$

Now recognize this as a geometric sum and precede.