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How to solve this limit?

$$\lim_{x\to1}\frac{x^2-1}{\ln(x)}=\frac00$$

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    Not about the question, but that was a troll edit. – 2017-01-14
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5 Answers 5

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Set $x-1=h\iff x=1+h$

$$\lim_{x\to1}\dfrac{x^2-1}{\ln x}=\dfrac{\lim_{h\to0}\dfrac{(h+1)^2-1}h}{\lim_{h\to0}\dfrac{\ln(1+h)}h}=?$$

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    :D That was pretty neat to see. – 2017-01-14
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    Very neat, but doesn't he still need L'Hôpital for the denominator? – 2017-01-14
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    @O.VonSeckendorff You need **something** to get past the $\ln$, whether it be known limits or series expansion. – 2017-01-14
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    @O.VonSeckendorff, This is an elementary formula of limit, readily available from $$\lim_{x\to0}\dfrac{e^x-1}x=1$$ See http://math.stackexchange.com/questions/690898/determine-the-following-limit-as-x-approaches-0-frac-ln1xx – 2017-01-14
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    @labbhattacharjee I didn't know/remember that one. Good to know. – 2017-01-14
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    @SimpleArt Well the series doesn't use L'Hôpital, as in your answer. I don't know how strict is OP with the requirement of no use of rule. – 2017-01-14
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    @SimpleArt Dr. MV just wrote a really slick proof using an inequality. – 2017-01-14
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    Happy New Year Lab! If the user is asking about this limit, then surely relying on knowledge of $\lim_{h\to 0}\frac{\log(1+h)}{h}=1$ is completely circular in my humble opinion. – 2017-01-14
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    @Dr.MV How to define $\ln(x)$ then? – 2017-01-14
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    @SimpleArt See my posted solution and follow the embedded link. – 2017-01-14
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In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1} \tag 1$$

for $x>0$.

Hence, we have from $(1)$

$$\frac{x^2-1}{x-1}\le \frac{x^2-1}{\log(x)}\le \frac{x(x^2-1)}{x-1} \tag 2$$

whence application of the squeeze theorem yields the coveted limit

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

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    Plz. Check (1). – 2017-01-14
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    @MyGlasses Thank you; I've edited. -Mark – 2017-01-14
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    Haha, nice... :-D – 2017-01-14
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    @SimpleArt Thank you. And if you follow the link, it will show a way forward that begins with the limit definition of the exponential function. – 2017-01-14
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With substitution $x=e^t$ we have: $$\lim_{x\to1}\frac{x^2-1}{\ln x}=\lim_{t\to0}\frac{e^{2t}-1}{t}=\lim_{t\to0}\frac{1+2t+\frac{(2t)^2}{2}-1}{t}=2$$ where $$e^\theta=1+\theta+\frac{\theta^2}{2}$$

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    We can employ $$\lim_{x\to0}\dfrac{e^x-1}x=1$$ – 2017-01-14
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    @labbhattacharjee from where? without proof? – 2017-01-14
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    Which identities can be used for limit ? http://math.stackexchange.com/questions/42679/proof-of-fx-ex-1-x-1-text-as-x-to-0-using-epsilon-delta-definiti OR http://2000clicks.com/MathHelp/CalculusLimitExponential.aspx – 2017-01-14
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    I mean, it is limit definition to derivative of $e^x$ at $x=0$. – 2017-01-14
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    @labbhattacharjee For using this limit, according to your links, _all_ apply l'hopital rule, derivative definition and Taylor expansion. My suppose is the part of Taylor expansion also, but I don't know where can find a proof without these techniques. – 2017-01-14
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Notice that

$$\ln(x)=(x-1)-\frac12(x-1)^2+\mathcal O(x^3)$$

and

$$x^2-1=(x-1)(x+1)$$

So the reciprocal of the limit is

$$\frac1L=\lim_{x\to1}\frac1{x+1}-\frac{(x-1)}{2(x+1)}+\mathcal O\left(\frac{(x-1)^2}{x+1}\right)=\frac12$$

So that

$$L=2$$

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By a shift of the variable your limit is equivalent to

$$\lim_{t\to0}\frac{t(t+1)}{\ln(1+t)}=\frac{\lim_{t\to0}t+1}{\lim_{t\to0}\dfrac{\ln(1+t)}t}=2.$$