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This is no homework, it's for exam practice.

Show that $\lim_{x\rightarrow 0}\frac{1}{x}ln(1+ax) = a$

where $a \in \mathbb{R}\setminus \left \{ 0 \right \}$ is chosen definitely / fixed (whatever this means...?) and $c = \frac{1}{a}$

Hint: Use the difference quotient.

So the difference quotient is defined like this:

$\lim_{x\rightarrow x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}= f'(x)$

But now, what is my $x_{0}$? I think $x_{0} = 0$.

The problem is if I try to use the difference quotient then it will not work for zero because I will have to divide through it since there is a $\frac{1}{x}$...

And using L'Hôpital won't work sadly because the criteria isn't fulfilled, we got $\frac{something}{0}$.

So how would I use the difference quotient on this and how do I know how I have to choose my $x_{0}$ if it hasn't been set by the teacher? I've always thought you choose a value where the function isn't defined, is that true?

I don't expect a solution but some answers to some of my questions would be very useful for me.

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Let $f(x)=\ln(1+ax)$ and $x_0=0$ in the difference quotient. Then $$\frac{f(x)-f(0)}{x-0}=\frac{\ln(1+ax)-\ln(1+a(0))}{x-0}=\frac{1}{x}\ln(1+ax).$$

This tells you that the limit you want is just $f'(0)$.


Proving the limit in this way (or using L'Hospital's rule as in the other answer) is dishonest/circular because you need to know the value of the limit to calculate $f'$.

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  • $\begingroup$ Thank you very much too for answer! How do you know that $x_{0} = 0$? Because the limes goes to zero? $\endgroup$ – cnmesr Jul 22 '16 at 14:09
  • $\begingroup$ Yes, because you have $\lim_{x\to 0}$ in the limit you want to prove. $\endgroup$ – smcc Jul 22 '16 at 14:13
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L'Hôpital actually works, as $\lim \log(1+ax) = \log(1+0) = 0$. Therefore $$\lim \frac{\log(1+ax)}{x} = \lim \frac{a}{1+ax} = a.$$

"$a$ is fixed" means that $a$ is a predetermined number, i.e., a constant.

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  • $\begingroup$ Ohhh you are totally right about that, thanks! For the fixed thing, am I allowed to assume something like: Let $a = 12$? $\endgroup$ – cnmesr Jul 22 '16 at 13:56
  • $\begingroup$ I would still like to see some answers to my other questions please, if anyone can I would be very happy to know :) $\endgroup$ – cnmesr Jul 22 '16 at 13:56
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    $\begingroup$ @cnmesr No. Your proof has to work for any choice of $a \neq 0$. $\endgroup$ – Alex Provost Jul 22 '16 at 13:57
  • $\begingroup$ Oh yeah that's right I forgot it needs to work for all :P $\endgroup$ – cnmesr Jul 22 '16 at 13:57

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