0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ (With $x^{1/17}$ instead of $x$.) $\endgroup$ – Hans Lundmark Oct 7 '14 at 19:38
  • $\begingroup$ 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. $\endgroup$ – Alan Oct 7 '14 at 19:45
11
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ This is definitely what I should have done since I was just working with the mean value theorem. $\endgroup$ – picaposo Oct 7 '14 at 19:52
  • $\begingroup$ The simplest. +1. $\endgroup$ – Did Oct 8 '14 at 5:37
7
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $a=(x+7)^\frac{1}{17},\;\; b=x^\frac{1}{17}, \;\;n=17.$

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.