Limit of over the real line

Question

I am trying to find $$ \lim_{x \to \infty} \biggl(\frac{x + \log9}{x - \log9} \biggr)^x $$

While I know the methods for getting limits of various indeterminate forms like like $\frac{0}{0}, \frac{\infty}{\infty},1^\infty,0\times0,\infty \times \infty$ this appears to be $\bigl(\frac{\infty}{\infty}\bigr)^\infty$.

How to convert it to some other indeterminate form whose limit can be found by standard methods.

are we doing $\biggl( \frac{1 + \frac{\log9}{x}}{1 - \frac{\log9}{x}} \biggr)$? — Vincent, Aug 09 '17 at 14:20
I will take the limit of the log of that function and use L'Hospital rule. — dh16, Aug 09 '17 at 14:27
If you want systematic computation of a wide class of limits, asymptotic expansions is what you need. For some motivation of the technique see this and for some examples see this and this and this. — user21820, Aug 09 '17 at 15:06

score 4 · Answer 1 · answered Aug 09 '17 at 14:46

By a shift of the variable

$$ \lim_{x \to \infty} \biggl(\frac{x + \log9}{x - \log9} \biggr)^x =\lim_{y \to \infty} \biggl(\frac{y +2 \log9}{y} \biggr)^{y+\log 9} =\lim_{y \to \infty} \biggl(1+\frac{2 \log9}{y} \biggr)^y\biggl(1+\frac{2 \log9}{y} \biggr)^{\log 9}.$$

The first factor is recognized as a case of the exponential function,

$$\lim_{y \to \infty} \biggl(1+\frac{2 \log9}{y} \biggr)^y=\lim_{z \to \infty} \biggl(1+\frac1z \biggr)^{z\,2 \log9}=e^{2\log 9}=\color{green}{81}$$ while the second factor tends to $1$.

score 3 · Accepted Answer · answered Aug 09 '17 at 14:35

Let $$f(x) = \left( \frac{x+\log 9}{x-\log 9} \right)^x.$$ We want to compute $\lim_{x\to\infty} \log f(x)$, and then exponentiate the result. $$ \lim_{x\to\infty} x \log \frac{x+\log 9}{x-\log 9} = \lim_{x\to\infty} \frac{\log \frac{x+\log 9}{x-\log 9}}{\frac{1}{x}} = \lim_{x\to\infty}\frac{(2\log 9) x^2}{x^2 - \log^2 9} = 2\log 9, $$ where the second equality follows from L'Hospital Rule. The final answer should be $e^{2\log 9} = 81$.

haqnatural · Answer 3 · 2017-08-09T14:51:40.483

1

this limit in the ${ 1 }^{ \infty }$ form so $$\lim _{ x\to \infty } \left( \frac { x+\log 9 }{ x-\log 9 } \right) ^{ x }=\lim _{ x\to \infty } \left( 1+\frac { 2\log 9 }{ x-\log 9 } \right) ^{ x }=\lim _{ x\to \infty }{ \left[ \left( 1+\frac { 2\log 9 }{ x-\log 9 } \right) ^{ \frac { x-\log 9 }{ 2\log 9 } } \right] } ^{ \frac { 2\log 9 }{ x-\log 9 } x }=\\ =\lim _{ x\to \infty }{ e } ^{ \frac { 2\log 9 }{ x-\log 9 } x }={ e }^{ 2\log 9 }=81$$

edited Aug 09 '17 at 14:51

answered Aug 09 '17 at 14:22

haqnatural

21,578

1

@Adayah Usually easy to do by squeeze theorem and/or Bernoulli's inequalities. – Simply Beautiful Art Aug 09 '17 at 14:24
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@SimplyBeautifulArt I deleted my comment when I noticed more mistakes so the author has time to correct them. In particular, the result is wrong. – Adayah Aug 09 '17 at 14:25
@Adayah Yes. It may help to observe that $$\left(1+\frac a{x-b}\right)^x\sim_\infty\left(1+\frac ax\right)^x$$ – Simply Beautiful Art Aug 09 '17 at 14:27
@Adayah,where is the mistake,i don't understand – haqnatural Aug 09 '17 at 14:30
The $x$ in the exponent disappeared. – Adayah Aug 09 '17 at 14:33
@haqnatural Aherm, the limit is probably $e^{2\log9}=81$ wink wink – Simply Beautiful Art Aug 09 '17 at 14:34
@haqnatural Well, you haven't corrected it yet, so perhaps I'll be more explicit. $$x\ne\frac{x-\log9}{2\log9}\frac{2\log9}{x-\log9}$$That is, you seem to be missing something by the third limit. – Simply Beautiful Art Aug 09 '17 at 14:49
Oh, sorry, guys, i was blind,I missed $x$,adn didn't see this comments,thank you – haqnatural Aug 09 '17 at 14:53

Marios Gretsas · Answer 4 · 2017-08-13T17:27:24.920

Write $$\biggl(\frac{x + \log9}{x - \log9} \biggr)^x =e^{x\ln{(\frac{x + \log9}{x - \log9})}}$$ I have not checked if the limit can be solved with this form but this is a form you know and you can try working with it.

If you want you can solve the limit $$\lim_{x \rightarrow +\infty}\ln{(\frac{x + \log9}{x - \log9})}$$

$P.S:$

This answer was downvoted before some minutes but i do not understand why,because the O.P did not ask for his limit to be solved completely.

Limit of over the real line

4 Answers4