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I have to find $$\lim_{x\to 0} \left( \frac{1}{\log({x+1})}-\frac{1}{\log({x+\sqrt{x^2+1}})}\right)$$

using only notable limits no derivatives I'm completely stuck, I tried 4 times, using logarithms property and notable limits and every time I come to a more complex situation. When I was close to the solution I figured out that my solution was totally wrong due to the incorrect use of notable limits. Please help me (I'm sorry for the numerous questions regarding calculus of limits but today I'm gonna find plenty of limits because tomorrow I'm going to have a test..) I managed to arrive to this form: lim (1/x-(2)/(x^2+2x)) using correctly the notable limits but I can't continue from here...

Dipok
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  • If you only want to find the limit try something like wolframalpha to get the result and maybe the result give you hints how to solve it. Furthermore please use $\LaTeX$ to make your limit legible – Dominic Michaelis Nov 24 '13 at 14:45
  • Obviously I first saw the result with wolfram alpha then I tried to solve this limit without any success... – Dipok Nov 24 '13 at 14:48
  • What are "notable limits"? – Daniel Fischer Nov 24 '13 at 14:50
  • @DanielFischer In Italy, fundamental limits such as $\lim_{x\to0}\frac{\sin x}{x}$, $\lim_{x\to0}\frac{e^x-1}{x}$ and similar are called limiti notevoli. – egreg Nov 24 '13 at 16:39
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    @egreg Thanks. But now the question is, what is included in "and similar"? I guess that depends on the course/lecturer? – Daniel Fischer Nov 24 '13 at 16:45
  • @DanielFischer Yes, the standard toolset is not fixed. I continue to believe that this is a wrong way to teach limits. – egreg Nov 24 '13 at 16:46
  • @egreg I think your belief is entirely correct. – Daniel Fischer Nov 24 '13 at 16:48
  • duplicate of http://math.stackexchange.com/questions/552661/proving-lim-limits-x-to0-left-frac1-logx-sqrt1x2-frac1-log1/ – Paramanand Singh Nov 26 '13 at 05:05
  • Regarding the use of "notable limits", I believe OP wants a procedure which follows standard limit theorems taught in a chapter on limits in an introductory calculus course. This obviously avoids techniques based on differentiation and integration. Not all limits can be solved by notable limits, but many can be. Notable limits include one limit each for algebraic func, trig func, logarithm and exponential. Two have been mentioned by egreg and remaining are $\lim_{x \to a}\dfrac{x^{n} - a^{n}}{x - a} = na^{n - 1}, \lim_{x \to 0}\dfrac{\log(1 + x)}{x} = 1$. – Paramanand Singh Nov 26 '13 at 05:15
  • All the four notable limits can be proved without techniques involving differentiation provided one chooses a suitable definition of the required functions. – Paramanand Singh Nov 26 '13 at 05:16
  • @Dipok Could you please try writing the addition using MathJax formatting? It's difficult to understand what the formula means. – egreg Nov 26 '13 at 13:05
  • The problem is that I don't know how to use it... – Dipok Nov 26 '13 at 13:17

3 Answers3

1

I know this isn't a solution along the requested lines, but it can give some ideas.

Set $t=\log(x+\sqrt{x^2+1})$ so that $x=\sinh x$ and the limit becomes $$ \lim_{t\to0}\left(\frac{1}{\log(1+\sinh t)}-\frac{1}{t}\right)= \lim_{t\to0}\frac{t-\log(1+\sinh t)}{t\log(1+\sinh t)} $$ Now $\log(1+u)=u-u^2/2+o(u^3)$ and $\sinh t=t+o(t^3)$ so $$ t-\log(1+\sinh t)=t-\sinh t+\frac{1}{2}\sinh^2t+o(t^3)= t-t+\frac{1}{2}t^2+o(t^3) $$ and therefore the requested limit is $\dfrac{1}{2}$.

(Note: checking with a calculator I get that the expression, for $x=0.001$, evaluates to $\approx0.49975$.)

egreg
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  • I fear that this is one of those limits that can be found with only notable limits... however the result is correct! – Dipok Nov 24 '13 at 16:57
1

If you have a notable limit of the form

$$\lim_{x\to 0} \left( \frac{1}{\log({x+1})}-\frac{1}{x}\right)=L$$

then you can conclude that

$$\begin{align} \lim_{x\to 0} \left( \frac{1}{\log({x+1})}-\frac{1}{\log({x+\sqrt{x^2+1}})}\right)&=\lim_{x\to 0} \left( \frac{1}{x}-\frac{1}{{x-1+\sqrt{x^2+1}}}\right)\\ &=\lim_{x\to 0} \left( \frac{1}{x}+\frac{x-1-\sqrt{x^2+1}}{2x}\right)\\ &=\lim_{x\to 0} \left( \frac{x+1-\sqrt{x^2+1}}{2x}\right)\\ &=\lim_{x\to 0} \left( \frac{1}{x+1+\sqrt{x^2+1}}\right)\\ &=\frac{1}{2} \end{align}$$

Note, the actual value of the notable limit, $L$, drops out right away, but as it happens, it's also equal to $1/2$.

Barry Cipra
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  • This is a great answer. It reduces the problem to a very simple algebraic calculation. However there is a catch that evaluation of $L$ requires use of L'Hospital rule or Taylor' series. A simpler approach using only "notable limits" is given at http://math.stackexchange.com/questions/552661/proving-lim-limits-x-to0-left-frac1-logx-sqrt1x2-frac1-log1/553878#553878 – Paramanand Singh Nov 26 '13 at 05:09
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Here is my solution: lim (1ln(x+1) - 1/ln(x+sqrt(x^2+1)) lim (1/(x)-1/(lnx+ln(1+sqrt(x^2+1)/x))) lim (1/(x)-1/(lnx+ln(1+x/2+1/x))) lim (1/(x)-1/(lnx+ln(2x+x^2+2)-ln(x)-ln2)) lim (1/(x)-1/(ln(1+x+(x^2)/2))) lim (1/(x)-1/(2x+x^2)) lim (1/(x+2))=1/2

Using only notable limits regarding natural logarithm, square root and exponential.

Dipok
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  • You have a mistake in the first step itself where you replace $1/\ln(1 + x)$ with $1/x$. This is not allowed by use of any notable limits or standard limit theorems. A proper solution to this problem using notable limits is already available at http://math.stackexchange.com/questions/552661/proving-lim-limits-x-to0-left-frac1-logx-sqrt1x2-frac1-log1/553878#553878 – Paramanand Singh Nov 26 '13 at 17:02
  • [focusing on the denominator you are referring] I have multiplied with x/x so that I have a notable limit that is 1 that multiplies the remaning x of the fraction that I have introduced. So the fraction becomes 1/x. I think this way is easier. – Dipok Nov 26 '13 at 17:50
  • I think i need to explain further. If $\lim_{x \to a} f(x) = A$ then it does not imply that $\lim_{x \to a}{f(x)g(x) + h(x)} = \lim_{x \to a}{Ag(x) + h(x)}$. Rather it should be $\lim_{x \to a}{f(x)g(x) + h(x)} = A\lim_{x \to a}g(x) + \lim_{x \to a}h(x)$. Unless you know more about $\lim_{x \to a}h(x)$ and $\lim_{x \to a}g(x)$ you can't do such replacement of a function with its limit. In this connection see "Misuse of Rules of Limits" in this blog post http://paramanands.blogspot.com/2013/11/teach-yourself-limits-in-8-hours-part-2.html – Paramanand Singh Nov 26 '13 at 18:16
  • Ok, unfortunately I'm not at that level, I'm attending secondary school and I only know that in calculus of a limit I can use algebric manipulation in order to use notable limits and solve the limit itself, indeed I found the limit in this case.(but I know that it doesn't work everytime). However thanks for your explanation! – Dipok Nov 26 '13 at 20:02
  • I understand that for beginners it is difficult to understand all the finer details (I was also in that phase long ago). But still one must keep track of "what is valid" and "what is not" while doing mathematical manipulations. And the reason your technique has worked in this question is explained in Barry Cipra's answer. You have been lucky where an invalid step also leads to a right answer. You need to be more careful while applying these manipulations. – Paramanand Singh Nov 27 '13 at 04:43
  • I'm aware of this, also my professor said me the same thing in a more generic sense, thank you! – Dipok Nov 27 '13 at 12:54