# Limit with natural log in the denominator: $\lim_{x\to1}{\frac{x^2 - 1}{\ln x}}$ [closed]

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

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

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

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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).

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

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!

• I love your "as was to be shown." It's very Euclid-esque. :D – Adam Hrankowski Sep 12 '15 at 23:27
• Very elegant proof, I might add. – Adam Hrankowski Sep 12 '15 at 23:36
• @AdamHrankowski Thank you!! I really appreciate the comment! – Mark Viola Sep 12 '15 at 23:52
• 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 $x \neq 1$) – user14972 Sep 13 '15 at 7:20
• Some links to post about the inequality used in your answer: math.stackexchange.com/questions/324345/… and math.stackexchange.com/questions/1161278/… – Martin Sleziak Sep 13 '15 at 9:17

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

• (+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 $\ln (1+y)$. – Silverfish Sep 13 '15 at 2:51
• You can simplify by using a taylor series with remainder rather than the full series: $\log(1+x) = x + x h(x)$, where $\lim_{x \to 0} h(x) = 0$. – user14972 Sep 13 '15 at 7:16

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.

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

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

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 $x\to1$

• 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. ${}\qquad{}$ – Michael Hardy Sep 12 '15 at 18:35
• @ Michael is it ok now – Display name Sep 12 '15 at 18:40
• 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. ${}\qquad{}$ – Michael Hardy Sep 12 '15 at 18:42