$\displaystyle \lim_{x \to 0}\frac{\ln(1+5x)}{x}$ 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.
 A: $$\lim_{h\to0}\frac{\ln(1+h)}h=1$$
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 $a$
A: $5\times \lim_{y\to 0}{\log(1+y)\over y}=5\times 1=5$ where $y=5x$
A: $ 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}+...$
Now,
$ \lim_{x\rightarrow 0}\dfrac{ln(1+5x)}{x}= \dfrac{5x-\dfrac{25x^2}{2}+\dfrac{125x^3}{3}-\dfrac{625x^4}{4}+...}{x}$
$\lim_{x\rightarrow 0}\dfrac{ln(1+5x)}{x}= \dfrac{5x}{x}-\dfrac{25x^2}{2x}+\dfrac{125x^3}{3x}-\dfrac{625x^4}{4x}+...$
$=\lim_{x\rightarrow 0} 5-\dfrac{25x}{2}+\dfrac{125x^2}{3}-\dfrac{625x^3}{4}+...$
$=5$
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.
$\int 1-x+x^2-x^3 ..) dx = x - \dfrac{x^2}{2} + \dfrac{x^3}{3}-\dfrac{x^4}{4}...$
A: $$\dfrac{\ln(1+5x)}{x} = \dfrac{\ln(1+5x)-\ln(1)}{x-0}$$
so this limit is just the definition of the derivative of $\ln(1+5x)$ at $x=0$.
A: Using Hospital's rule, we have that $$\lim_{x\to 0} \frac{ln(1+5x)}{x}=\lim _{x\to 0} \frac{(ln(1+5x))'}{(x)'}=\lim_{x\to 0} \frac{5}{1+5x}=5$$
