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So have I have to find this limit: $$\lim_{x \to 0}\frac{\ln(1+5x)}{x}$$

With substitution of $c$ into the function, I get $\lim = \frac{0}{0}$ which is undefined. If this were, say, a polynomial, I'd try to factor it out, but in this case I just don't see a strategy here.

How should I go about this? Any links for extra reading would be especially handy.

Restriction: I can't use the L'Hospital rule.

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are you familiar with L hospital rule? – Praphulla Koushik Dec 9 '13 at 15:57
Do you know L'hopital's rule? – Mhenni Benghorbal Dec 9 '13 at 15:58
@Iota : If you think L hospitel will lead to circular reasoning, it is advisable to look below for Mitsos answer... – Praphulla Koushik Dec 9 '13 at 16:01
I do have that one restriction, unfortunately. – Morgan Wilde Dec 9 '13 at 16:03
@MorganWilde : Very good... you said this restriction just after getting so much excited about that... You could have said this before:( – Praphulla Koushik Dec 9 '13 at 16:04
up vote 4 down vote accepted


Observe that $(i)$the limit variable, $(ii)$the denominator & $(iii)$the argument inside the parenthesis of logarithm must be same

Also, if $h\to0, a\cdot h\to0$ for finite $a$

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This seems the most helpful answer. I wonder how should I proceed, given I have $1+5x$ inside the natural log, and just $x$ in the denominator? – Morgan Wilde Dec 9 '13 at 16:08
OK, I see that now, multiply by $1$ in the form of $\frac{5}{5}$. Thanks! – Morgan Wilde Dec 9 '13 at 16:09
@MorganWilde, set $h=5x$ – lab bhattacharjee Dec 9 '13 at 16:09

$$\dfrac{\ln(1+5x)}{x} = \dfrac{\ln(1+5x)-\ln(1)}{x-0}$$ so this limit is just the definition of the derivative of $\ln(1+5x)$ at $x=0$.

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just a query : $$\frac{d(\ln x)}{dx}=\frac1x$$ comes from the fact that $$\lim_{h\to0}\frac{\ln(1+h)}h=1$$, So, is it legal to use the derivative in calculating the limit? – lab bhattacharjee Dec 10 '13 at 8:37
It depends on how you define the logarithm and exponential functions. If you define $\ln$ as the inverse function of $\exp$, then $\dfrac{d}{dx} \ln(x) = \dfrac{1}{x}$ is a consequence of $\dfrac{d}{dt} \exp(t) = \exp(t)$. If you define $\ln(x)$ as $\int_1^x \dfrac{dt}{t}$, then ... – Robert Israel Dec 12 '13 at 9:23

$5\times \lim_{y\to 0}{\log(1+y)\over y}=5\times 1=5$ where $y=5x$

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$ ln(1+x)= x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+...$

Which for your case leads to

$ ln(1+5x)= 5x-\dfrac{25x^2}{2}+\dfrac{125x^3}{3}-\dfrac{625x^4}{4}+...$


$ \lim_{x\rightarrow 0}\dfrac{ln(1+5x)}{x}= \dfrac{5x-\dfrac{25x^2}{2}+\dfrac{125x^3}{3}-\dfrac{625x^4}{4}+...}{x}$

$\lim_{x\rightarrow 0}\dfrac{ln(1+5x)}{x}= \dfrac{5x}{x}-\dfrac{25x^2}{2x}+\dfrac{125x^3}{3x}-\dfrac{625x^4}{4x}+...$

$=\lim_{x\rightarrow 0} 5-\dfrac{25x}{2}+\dfrac{125x^2}{3}-\dfrac{625x^3}{4}+...$


Edit: how does the series form come about, one (loose) way.

$ \dfrac{d\ ln(1+x)}{dx}=\dfrac{1}{1+x}$

Expanding it in a series form.

$ \dfrac{d\ ln(1+x)}{dx}=\dfrac{1}{1+x} = 1-x+x^2-x^3 ..$

Now to get $ ln(1+x)$ from $\dfrac{d\ ln(1+x)}{dx}$ , integrate the series term wise. $\int 1-x+x^2-x^3 ..) dx = x - \dfrac{x^2}{2} + \dfrac{x^3}{3}-\dfrac{x^4}{4}...$

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Which property of $log_e$ did you use to deduce that $ln(1+x) = x - \frac{x^2}{2} + ..$? – Morgan Wilde Dec 9 '13 at 20:38
@MorganWilde Excellent question. Well $\dfrac{d ln(1+x)}{dx}=\dfrac{1}{1+x}$ Lets take a series form $\dfrac{1}{1+x}=1-x+x^2$. Now this is $\dfrac{d ln(1+x)}{dx}$. So to get $ln(1+x)$ Integrate termwise which leads to the expression. – user3077268 Dec 9 '13 at 20:40
I would very much like to hear the answer to it, if it wouldn't bother you too much. – Morgan Wilde Dec 9 '13 at 20:41
@MorganWilde Excellent question. Well $\dfrac{d\ ln(1+x)}{dx}=\dfrac{1}{1+x}$ Lets take a series form $\dfrac{1}{1+x}=1-x+x^2$. Now this is $\dfrac{d\ ln(1+x)}{dx}$. So to get $ln(1+x)$ Integrate termwise which leads to the expression. Now think what will happen if you do $ln(1+x) + ln(1-x)$ and $ln(1+x)-ln(1-x)$. You would get beautiful infinite series containing only odd or even powers. – user3077268 Dec 9 '13 at 20:50
Please add this to your answer, since it is a very valuable addition. Thanks a lot! – Morgan Wilde Dec 9 '13 at 20:50

Using Hospital's rule, we have that $$\lim_{x\to 0} \frac{ln(1+5x)}{x}=\lim _{x\to 0} \frac{(ln(1+5x))'}{(x)'}=\lim_{x\to 0} \frac{5}{1+5x}=5$$

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I assume you mean as $x\to 0$. – Sam DeHority Dec 9 '13 at 16:02
@SamDeHority correct! – Haha Dec 9 '13 at 16:16

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