6
$\begingroup$

Value of $\displaystyle\lim_{x\to1}{\frac{x^2 - 1}{\ln x}}$

The answer is given to be $2$. I'd appreciate an explanation.

$\endgroup$

closed as off-topic by user223391, saz, user99914, user133281, drhab Sep 13 '15 at 12:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, saz, Community, user133281, drhab
If this question can be reworded to fit the rules in the help center, please edit the question.

8
$\begingroup$

Since simple substitution of $x:=1$ would yield the indeterminate form $\frac{0}{0}$,

L'Hôpital's rule to the rescue:

$$\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 1}\frac{f'(x)}{g'(x)}$$

So, take the derivative of the top and the bottom (not the derivative of the top divided by the bottom).

$$\lim_{x\rightarrow 1}\frac{x^2-1}{\ln x} = \lim_{x\rightarrow 1}\frac{2x}{1/x}=\lim_{x\rightarrow 1}2x^2= 2$$

$\endgroup$
5
$\begingroup$

We don't need to rely on L'Hospital's Rule (not that there's anything wrong with using it here).

We only need recall that $\frac{x-1}{x}\le \log x \le x-1$ for $x>0$, with strict inequalities for $x\ne 1$.

Then, for $x\ne 1$

$$x+1=\frac{x^2-1}{x-1}<\frac{x^2-1}{\log x}<\frac{x^2-1}{\frac{x-1}{x}}=x(x+1)$$

By the squeeze theorem, we have

$$\lim_{x\to 1}\frac{x^2-1}{\log x}=2$$

as was to be shown!

$\endgroup$
  • $\begingroup$ I love your "as was to be shown." It's very Euclid-esque. :D $\endgroup$ – Adam Hrankowski Sep 12 '15 at 23:27
  • $\begingroup$ Very elegant proof, I might add. $\endgroup$ – Adam Hrankowski Sep 12 '15 at 23:36
  • $\begingroup$ @AdamHrankowski Thank you!! I really appreciate the comment! $\endgroup$ – Mark Viola Sep 12 '15 at 23:52
  • $\begingroup$ Those inequalities should be $\leq$ rather than $<$, since all three terms are equal at $x=1$. (or clarify that they were only meant to hold for $x \neq 1$) $\endgroup$ – user14972 Sep 13 '15 at 7:20
  • 2
    $\begingroup$ Some links to post about the inequality used in your answer: math.stackexchange.com/questions/324345/… and math.stackexchange.com/questions/1161278/… $\endgroup$ – Martin Sleziak Sep 13 '15 at 9:17
3
$\begingroup$

They told us not to use l'Hospital if not 100% sure.
So for beginners with a little bit of stamina this is a nice method.
You can directly reduce the diverging factor here.
Recall $$\ln(x) = \ln ((x-1)+1) = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}(x-1)^n,$$ therefore $$\lim _{ x\rightarrow 1 } \frac { x ^2 -1 }{ \ln ( x) }=\lim _{ x\rightarrow 1 }{ \frac{ (x+1)(x-1) }{ \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}(x-1)^n } } =\lim _{ x\rightarrow 1 }{ \frac{ (x+1)}{ \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}(x-1)^{n-1} } } . $$

By shifting the index we get $$ \lim _{ x\rightarrow 1 }{ \frac{ (x+1)}{ \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}(x-1)^{n-1} } } = \lim _{ x\rightarrow 1 }{ \frac{ (x+1)}{ \sum_{i=0}^\infty\frac{(-1)^{i+2}}{i+1}(x-1)^{i} } } .$$ Since

$$ \lim _{ x\rightarrow 1 }{ \sum_{i=0}^\infty\frac{(-1)^{i+2}}{i+1}(x-1)^{i} } = 1,$$ we have $$\lim _{ x\rightarrow 1 }{ \frac{ (x+1)}{ \sum_{i=0}^\infty\frac{(-1)^{i+2}}{i+1}(x-1)^{i} } } ={ \frac{\lim _{ x\rightarrow 1 } (x+1)}{\lim _{ x\rightarrow 1 } \sum_{i=0}^\infty\frac{(-1)^{i+2}}{i+1}(x-1)^{i} } } =\frac{2}{1}=2.$$

$\endgroup$
  • 2
    $\begingroup$ (+1) I think series solutions are often much more insightful about the behavior of a function than L'Hopital is. Maybe easier to let $y=x-1$, and take the limit as $y \to 0$. Then you can just use the expansion of $\ln (1+y)$. $\endgroup$ – Silverfish Sep 13 '15 at 2:51
  • $\begingroup$ You can simplify by using a taylor series with remainder rather than the full series: $\log(1+x) = x + x h(x)$, where $\lim_{x \to 0} h(x) = 0$. $\endgroup$ – user14972 Sep 13 '15 at 7:16
2
$\begingroup$

Hint : $$\frac{x^2-1}{\log x} = \frac{(x-1)(x+1)}{\log x}= \frac{x+1}{\frac{\log x}{x-1}}$$

And recall this.

$\endgroup$
2
$\begingroup$

$$\lim _{ x\rightarrow 1 }{ \frac { { x }^{ 2 }-1 }{ \ln { x } } =\lim _{ x\rightarrow 1 }{ \frac { x+1 }{ \frac { \ln { x } }{ x-1 } } } } =\frac { \lim _{ x\rightarrow 1 }{ \left( x+1 \right) } }{ \lim _{ x\rightarrow 1 }{ \left( \frac { \ln { x } }{ x-1 } \right) } } =2$$

$\endgroup$
  • $\begingroup$ Is there a simple proof that $\lim_{x \to 1}\ln(x)/(x-1)=1$ or are you relying on l'hopital $\endgroup$ – ASKASK Sep 12 '15 at 20:17
  • $\begingroup$ Your have 0/0 in the denominator. $\endgroup$ – Adam Hrankowski Sep 12 '15 at 20:30
  • 1
    $\begingroup$ $$\lim _{ x\rightarrow 1 }{ \frac { \ln { x } }{ x-1 } } \\ t=\ln { x\quad \Rightarrow x={ e }^{ t } } \\ x\rightarrow 1,t\rightarrow 0\\ \lim _{ t\rightarrow 0 }{ \frac { t }{ { e }^{ t }-1 } =\lim _{ t\rightarrow 0 }{ \frac { 1 }{ \frac { { e }^{ t }-1 }{ t } } } =1 } $$ and see here :math.stackexchange.com/questions/42679/… $\endgroup$ – haqnatural Sep 13 '15 at 6:03
2
$\begingroup$

As the function is of form $\frac{0}{0}$ we can apply LHR

therefore the derivative of function becomes $\frac{2x}{\frac{1}{x}}$

which is equal to

$2x^2$

now you know $x\to1$

$\endgroup$
  • $\begingroup$ It seems misleading to say "the function becomes" this other expression. It makes it sound as if it's the same function, rather than a function with the same limit. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 12 '15 at 18:35
  • $\begingroup$ @ Michael is it ok now $\endgroup$ – Display name Sep 12 '15 at 18:40
  • $\begingroup$ The newly edited version seems misleading in a different way. "the derivative becomes${}\,\ldots$ makes it sound as if that is $\dfrac d {dx}\, \dfrac{x^2-1}{\log x}$. But what you've done is to take the derivative of the numerator and the denominator separately, not the derivative of the quotient. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 12 '15 at 18:42

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