How to solve this limit?
$$\lim_{x\to1}\frac{x^2-1}{\ln(x)}=\frac00$$
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Welcome to the site! Please see the MathJax guide to see how to format your math. – Simply Beautiful Art Jan 14 '17 at 14:48
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Not about the question, but that was a troll edit. – OFRBG Jan 14 '17 at 14:49
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Also, please do not mess up the pretty $\LaTeX$... – Simply Beautiful Art Jan 14 '17 at 14:50
5 Answers
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Set $x-1=h\iff x=1+h$
$$\lim_{x\to1}\dfrac{x^2-1}{\ln x}=\dfrac{\lim_{h\to0}\dfrac{(h+1)^2-1}h}{\lim_{h\to0}\dfrac{\ln(1+h)}h}=?$$
lab bhattacharjee
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@O.VonSeckendorff You need something to get past the $\ln$, whether it be known limits or series expansion. – Simply Beautiful Art Jan 14 '17 at 14:56
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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 – lab bhattacharjee Jan 14 '17 at 14:56
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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. – OFRBG Jan 14 '17 at 14:58
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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. – Mark Viola Jan 14 '17 at 15:05
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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
$$\lim_{x\to 1}\frac{x^2-1}{\log(x)}=2$$
Mark Viola
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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. – Mark Viola Jan 14 '17 at 15:11
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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 $$e^\theta=1+\theta+\frac{\theta^2}{2}$$
Nosrati
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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 – lab bhattacharjee Jan 14 '17 at 15:04
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I mean, it is limit definition to derivative of $e^x$ at $x=0$. – Simply Beautiful Art Jan 14 '17 at 15:09
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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. – Nosrati Jan 14 '17 at 15:09
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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
$$L=2$$
Simply Beautiful Art
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By a shift of the variable your limit is equivalent to
$$\lim_{t\to0}\frac{t(t+1)}{\ln(1+t)}=\frac{\lim_{t\to0}t+1}{\lim_{t\to0}\dfrac{\ln(1+t)}t}=2.$$