Evaluate limit of the type : $\infty - \infty$?

Question

Evaluate:

$$\lim_{x\to+\infty}(x+7)^\frac{1}{17}-x^\frac{1}{17}$$

I don't know where to begin to be honest. I know that when a limit evaluates to:

$$\infty-\infty$$

it simply means that it the limit is still indeterminate and could be anything. I cannot see a way to rewrite the limit so that it can be resolved.

possible duplicate of Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ — Hans Lundmark, Oct 07 '14 at 19:36
the usual trick here is to turn something where we are subtracting two infinities into something where we are dividing two infinities. So, what you do is you multiply the top and bottom by the "conjugate" of the top, since multiplying anything by something of the form $\frac a a=1$ doesnt change it. I'm not sure of the proper fraction for this equation to simplify the algebra, hopefully someone is, but its a general technique. — Alan, Oct 07 '14 at 19:45

score 11 · Accepted Answer · answered Oct 07 '14 at 19:38

11

Use the mean value theorem: $$ (x+7)^{1/17}-x^{1/17}=\frac{1}{17}\,(\xi_x)^{-16/17}\cdot7 $$ with $x\le \xi_x\le x+7$.

answered Oct 07 '14 at 19:38

Julián Aguirre

76,354

2

This is definitely what I should have done since I was just working with the mean value theorem. – picaposo Oct 07 '14 at 19:52
The simplest. +1. – Did Oct 08 '14 at 05:37

score 7 · Answer 2 · answered Oct 07 '14 at 19:35

7

$$(x+c)^\alpha=x^\alpha\cdot(1+c/x)^\alpha=x^\alpha(1+c\alpha/x+o(1/x))=x^\alpha+c\alpha x^{\alpha-1}+o(x^{\alpha-1})$$

answered Oct 07 '14 at 19:35

Did

279,727

score 0 · Answer 3 · answered Oct 07 '14 at 19:46

0

Hint:
Use the identity $$a^n-b^n={(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+ \ldots +a b^{n-2}+b^{n-1})}$$ with $a=(x+7)^\frac{1}{17},\;\; b=x^\frac{1}{17}, \;\;n=17.$

answered Oct 07 '14 at 19:46

M. Strochyk

8,397

score 0 · Answer 4 · answered Oct 07 '14 at 19:52

Using the Hint of Hans Lundmark, let $u=\sqrt[17]{x}$ and rewrite the limit as:

$$\lim_{u\to\infty}\sqrt[17]{u^{17}+1}-u$$

Then, $$\lim_{u\to\infty}\frac{\sqrt[17]{1+\frac{1}{u^{17}}}-1}{\frac{1}{u}}$$

Now, substitute again to $y=\frac{1}{u}$, so $u\to\infty\Rightarrow y\to0$ and $$\lim_{y\to0}\frac{\sqrt[17]{1+y^{17}}-1}{y}$$

Using L'Hopital, $$\lim_{y\to0}\frac{\sqrt[17]{1+y^{17}}-1}{y}=\lim_{y\to0}\frac{\frac{1}{17\cdot\sqrt[17]{1+y^{17}}^{16}}\cdot17y^{16}}{1}=\lim_{y\to0}\frac{y^{16}}{\sqrt[17]{1+y^{17}}^{16}}=0$$

Evaluate limit of the type : $\infty - \infty$?

4 Answers4