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$$\lim_{x \to 0^+} \frac{x}{\ln (5x + 1)} = {1 \over 5}$$

First, What I tried to do is dividing by $x$, but it didn't work out.
By the way, It's a common approach to find a limit. Why is it failing here?

Second, I played with the logarithm rules which only brought me to:
$$\log _{5x + 1} e^x = x \log_{5x + 1} e$$

It might be a starter, but I don't know what to do from this point.

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    $\begingroup$ Please notice my edits to your question. An expression like \mathop{lim}\limits_{x \to {0^+}} is an abominable way of writing MathJax or TeX code. I changed it to \lim_{x \to 0^+}. I also deleted some superfluous curly braces. Those can make editing harder. $\endgroup$ Dec 15 '13 at 22:04
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    $\begingroup$ You should know that $\lim \limits_{y\to 0^+}\left(\dfrac{\log (y+1)}{y}\right)=1$. Hint: $\lim \limits_{x\to 0^+}\left(\dfrac x{\log(5x+1)}\right)=\dfrac 1 5\lim \limits_{x\to 0^+}\left(\dfrac {5x}{\log(5x+1)}\right)$. $\endgroup$
    – Git Gud
    Dec 15 '13 at 22:09
  • $\begingroup$ @GitGud I'd accept your solution if you wrote it as an answer. The reason is I cannot use l'hospital rule. I should have mentioned it. Thanks! $\endgroup$
    – AndrePoole
    Dec 16 '13 at 9:20
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Since both denominator and numerator approaches 0 at the limit of $x\rightarrow 0^+$, you can use the l'hospital rule to find this limit

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Use L'Hospital's Rule

For indeterminate forms, like what you have $$\lim_{x\to0^+}\frac{x}{\ln(5x+1)} = \frac{0}{0}$$ And given that $\lim_{x\to0}\frac{f(x)}{g(x)}, \space f'(x) \space g'(x)$ exists.

L'Hospital's Rule says: $$\lim_{x\to0^+}\frac{x}{\ln(5x+1)}= \lim_{x\to0^+}\frac{(x)'}{(\ln(5x+1))'}=...$$

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Take a look at L'Hopital's rule. When the numerator and denominator both go to 0 or infinity, you can take the derivative of both and then try finding the limit again.

Here's a good reference: http://www.youtube.com/watch?v=PdSzruR5OeE

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Or using the fact that $\ln(1+t)\sim_0t$, for $t=5x$.

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