Which is the solution of this limit? I have to calculate the following limit:
$$ \lim_{h\to 0} \frac{\frac{ln(1+h^4)}{h^4}-1}{h} $$
First of all I've put the same denominator and I obtain $$ \lim_{h\to 0} \frac{ln(1+h^4)-h^4}{h^5} $$
Then, I've separated it in two limits and I've used the equivalent infinitesim of $ ln(1+h^4) $ which is $ h^4 $
So now I've got $$ \lim_{h\to 0} \frac{1}{h}-\lim_{h\to 0} \frac{1}{h} $$
Each of those limits solution is infinite but if I put those two limits in one and I rest, the limit of 0 is equal to 0. So my question is:
The solution of this limit is $ 0 $ or $\infty$?
 A: Do a series expansion.  For $|x|<1$ and therefore in a neighborhood of $0$,
$$\log (1+x) = x - \frac{x^2}{2} + O(x^3).$$  Hence $$\frac{\log(1+x^4)}{x^4} = 1 - \frac{x^4}{2} + O(x^8)$$ and $$\frac{\frac{\log(1+x^4)}{x^4} - 1}{x} = -\frac{x^3}{2} + O(x^7).$$  As $x \to 0$, what happens?
A: The way you've split the limit is incorrect. You can write
$$\lim_{x\to a}(f(x) - g(x)) = \lim_{x\to a} f(x) - \lim_{x\to a}g(x),$$
if you know a priori that both the limits on the RHS do exist.
Otherwise, the RHS won't make sense.  

As for your actual limit, it is $0$. You can use the expansion 
$$\ln(1 + x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots$$
to conclude that or use L'Hosiptal.
If you use L'H, the limit simplifies as:
$$\lim_{h\to0}\dfrac{\frac{4h^3}{1 + h^4} - 4h^3}{5h^4} = \dfrac{4}{5}\lim_{h\to0}\dfrac{-h^4}{h(1 + h)} = \dfrac{4}{5}\lim_{h\to0}\dfrac{-h^3}{1 + h} = 0.$$
A: Try to do the substitution $h^4 = t$. This simplifies the problem quite a bit.
$$\lim_{t \to 0}\frac{\ln(t+1)-t}{t^{\frac 54}} \\ 
= \lim_{t \to 0}\frac{t-\frac{t^2}{2}+..-t}{t^{\frac 54}} \\
= \lim_{t \to 0} t^{\frac 34} + O(t^{\frac 74}) \\ 
= 0$$
